Worked Examples on Limits
A collection of fully worked examples involving the limit of a function.
| Exercise | $$ \lim_{x \rightarrow 3} \frac{x^2-3x}{x-3} $$ |
| Exercise | $$ \lim_{x \rightarrow 2} \frac{x^2-4}{x-2} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{\sqrt{x+4}-2}{3x} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{\sqrt{x-4}-2}{3x} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{x^3-x^2+4x}{x^5-x}$$ |
| Exercise | $$ \lim_{x \rightarrow 9} \frac{3-\sqrt{x}}{9-x} $$ |
| Exercise | $$ \lim_{x \rightarrow -\infty} 2x - \sqrt{4x^2+x} $$ |
| Exercise | $$ \lim_{x \rightarrow +\infty} \sqrt{x^2+3x-4}-x$$ |
| Exercise | $$ \lim_{x \rightarrow -5} \frac{x^2 + 3x -10}{x^2-25} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{\log \sqrt{x+1}}{x} $$ |
| Exercise | $$ \lim_{x \rightarrow +\infty} x+2-\sqrt{x^2+4x+8} $$ |
| Exercise | $$ \lim_{x \rightarrow -1} \frac{3x^2+x-10}{x^2-5x-14} $$ |
| Exercise | $$ \lim_{x \rightarrow 2} \frac{\sqrt{x^2-2x+9}-3}{x^3-x^2-x-2} $$ |
| Exercise | $$ \lim_{x \rightarrow -1} \frac{\sqrt{x^2+3}-2}{3-\sqrt{8-x^3}} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{\sin(x)}{x} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{\sin(3x)}{x} $$ |
| Exercise | $$ \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} $$ |
| Exercise | $$ \lim_{x \to 0} \frac{1 - \cos(3x)}{x^2} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{\sin(x)-x}{x^3} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^2} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{\cos(x)-1}{x} $$ |
| Exercise | $$ \lim_{x \rightarrow \frac{\pi}{4}} \frac{\tan(x)-1}{x - \frac{\pi}{4}} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{\sin x}{e^x-1} $$ |
| Exercise | $$ \lim_{x \rightarrow 2} \frac{\sqrt{x^2+5} - \sqrt{x^3+1}}{2x-x^2} $$ |
| Exercise | $$ \lim_{x \rightarrow 0 } \frac{3-\sqrt{9-x^2}}{2 \sqrt{1+x^2}-\sqrt{4+x^2}} $$ |
| Exercise | $$ \lim_{x \rightarrow 0} \frac{(e^{3x}-1) \sin x^4}{x^2 \log (1+x^3)} $$ |
| Exercise | $$ \lim_{x \to \infty} \frac{x + \sin(x)}{x - \cos(x)} $$ |
Exercise 1
We’re asked to evaluate the limit of the rational function as \( x \) approaches 3:
$$ \lim_{x \rightarrow 3} \frac{x^2 - 3x}{x - 3} $$
The function is undefined at \( x = 3 \), so the domain is \( (-\infty, 3) \cup (3, +\infty) \).
Although \( x = 3 \) is not part of the domain, it is a limit point - which means the limit might still exist.
Substituting directly gives an indeterminate form of type 0/0:
$$ \lim_{x \rightarrow 3} \frac{x^2 - 3x}{x - 3} = \frac{0}{0} $$
To resolve this, factor the numerator:
$$ \frac{x(x - 3)}{x - 3} $$
We can cancel out the common factor \( (x - 3) \) in the numerator and denominator:
$$ \lim_{x \rightarrow 3} x $$
This limit is straightforward to compute:
$$ \lim_{x \rightarrow 3} x = 3 $$
So, the function approaches the value 3 as \( x \) tends to 3.
Exercise 2
Let’s now compute the limit of the following function as \( x \) approaches 2:
$$ \lim_{x \rightarrow 2} \frac{x^2 - 4}{x - 2} $$
The function is undefined at \( x = 2 \), so its domain is \( (-\infty, 2) \cup (2, +\infty) \).
Again, since 2 is a limit point, we can attempt to evaluate the limit.
Direct substitution yields another indeterminate form of type 0/0:
$$ \lim_{x \rightarrow 2} \frac{x^2 - 4}{x - 2} = \frac{0}{0} $$
We factor the numerator using the difference of squares:
$$ \frac{(x - 2)(x + 2)}{x - 2} $$
Then we cancel the \( (x - 2) \) terms:
$$ \lim_{x \rightarrow 2} (x + 2) $$
Now the expression is continuous at \( x = 2 \), so we substitute directly:
$$ \lim_{x \rightarrow 2} (x + 2) = 4 $$
Therefore, the limit exists and equals 4.
Exercise 3
We’re asked to evaluate the following limit as \( x \) approaches zero:
$$ \lim_{x \rightarrow 0} \frac{\sqrt{x + 4} - 2}{3x} $$
We begin by identifying the domain of the function. The square root requires the expression inside it to be non-negative, so the function is defined on:
$$ D_f = [-4, 0) \cup (0, +\infty) $$
We now check the behavior of the limit as \( x \to 0 \). Direct substitution gives:
$$ \frac{\sqrt{4} - 2}{0} = \frac{0}{0} $$
We encounter an indeterminate form of type 0/0, so algebraic manipulation is required to proceed.
The standard approach here is to multiply the expression by the conjugate of the numerator:
$$ \lim_{x \rightarrow 0} \frac{\sqrt{x + 4} - 2}{3x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} $$
This transforms the numerator into a difference of squares:
$$ \lim_{x \rightarrow 0} \frac{(x + 4) - 4}{3x(\sqrt{x + 4} + 2)} = \lim_{x \rightarrow 0} \frac{x}{3x(\sqrt{x + 4} + 2)} $$
Canceling the common factor of \( x \) gives:
$$ \lim_{x \rightarrow 0} \frac{1}{3(\sqrt{x + 4} + 2)} $$
This expression is now continuous at \( x = 0 \), so we substitute directly:
$$ \frac{1}{3(\sqrt{4} + 2)} = \frac{1}{3(2 + 2)} = \frac{1}{12} $$
Conclusion: the value of the limit is \( \frac{1}{12} \).
Note. This limit can also be evaluated using L’Hôpital’s Rule, since it’s of the indeterminate form \( \frac{0}{0} \). Differentiating the numerator and denominator separately yields: $$ \lim_{x \rightarrow 0} \frac{\frac{1}{2\sqrt{x + 4}}}{3} = \frac{1}{3 \cdot 2\sqrt{4}} = \frac{1}{12} $$ The result matches the previous method, as expected.
Exercise 4
Consider the limit:
$$ \lim_{x \rightarrow 0} \frac{\sqrt{x - 4} - 2}{3x} $$
We begin with the domain analysis. The square root function is only defined for values where the argument is non-negative:
$$ x - 4 \geq 0 \quad \Rightarrow \quad x \geq 4 $$
Thus, the domain of the function is:
$$ D_f = [4, +\infty) $$
Since the point \( x = 0 \) lies outside the domain - and there is no neighborhood around 0 where the function is defined - the limit does not exist.
Conclusion: the limit is undefined due to domain restrictions.
Note. This simple example illustrates a fundamental principle: always analyze the domain of a function before attempting to evaluate a limit. Neglecting this step can lead to wasted effort on meaningless algebra. Often, the domain alone reveals the answer.
Exercise 5
We are tasked with evaluating the limit:
$$ \lim_{x \rightarrow 0} \frac{x^3 - x^2 + 4x}{x^5 - x} $$
Here, \( x = 0 \) is a point of accumulation for the domain but is excluded from it, since the denominator is zero at that point.
Substituting directly results in an indeterminate form of type 0/0:
$$ \frac{0 - 0 + 0}{0 - 0} = \frac{0}{0} $$
We proceed by factoring both the numerator and the denominator:
$$ \frac{x(x^2 - x + 4)}{x(x^4 - 1)} $$
Canceling the common factor of \( x \):
$$ \lim_{x \rightarrow 0} \frac{x^2 - x + 4}{x^4 - 1} $$
This expression is now well-defined at \( x = 0 \). Substituting gives:
$$ \frac{0 - 0 + 4}{0 - 1} = \frac{4}{-1} = -4 $$
Conclusion: the value of the limit is \( -4 \).
Exercise 6
In this exercise, we’re asked to compute the following limit:
$$ \lim_{x \rightarrow 9} \frac{3 - \sqrt{x}}{9 - x} $$
The domain of the function is:
$$ D_f = [0, 9) \cup (9, +\infty) $$
Explanation. The square root requires a non-negative argument, while the denominator vanishes at \( x = 9 \). Therefore, \( x = 9 \) is excluded from the domain.
However, since \( x = 9 \) is a limit point, the limit can still be evaluated.
Direct substitution yields the indeterminate form:
$$ \frac{3 - 3}{9 - 9} = \frac{0}{0} $$
This is a classic indeterminate form of type 0/0.
To resolve it, we rationalize the numerator using its conjugate:
$$ \lim_{x \rightarrow 9} \frac{3 - \sqrt{x}}{9 - x} \cdot \frac{3 + \sqrt{x}}{3 + \sqrt{x}} $$
This gives:
$$ \lim_{x \rightarrow 9} \frac{(3 - \sqrt{x})(3 + \sqrt{x})}{(9 - x)(3 + \sqrt{x})} $$
Applying the identity \( (a - b)(a + b) = a^2 - b^2 \):
$$ \lim_{x \rightarrow 9} \frac{9 - x}{(9 - x)(3 + \sqrt{x})} $$
Simplifying:
$$ \lim_{x \rightarrow 9} \frac{1}{3 + \sqrt{x}} $$
Now we can evaluate the limit directly:
$$ \lim_{x \rightarrow 9} \frac{1}{3 + \sqrt{x}} = \frac{1}{6} $$
Result: the value of the limit is \( \frac{1}{6} \).
Note. We could also apply L’Hôpital’s Rule: $$ \lim_{x \rightarrow 9} \frac{3 - \sqrt{x}}{9 - x} = \frac{0}{0} $$ Differentiate numerator and denominator: $$ \lim_{x \rightarrow 9} \frac{-\frac{1}{2\sqrt{x}}}{-1} = \frac{1}{2\sqrt{9}} = \frac{1}{6} $$ Same result, as expected.
Exercise 7
Now let’s evaluate the behavior of the function as \( x \to -\infty \):
$$ \lim_{x \rightarrow -\infty} 2x - \sqrt{4x^2 + x} $$
This limit may seem ambiguous at first.
As \( x \to -\infty \), the term \( 2x \) clearly diverges to \( -\infty \), while the square root involves a subtle \( \infty - \infty \) structure due to the competing terms \( 4x^2 \to +\infty \) and \( x \to -\infty \):
$$ \lim_{x \rightarrow -\infty} 2x - \sqrt{4x^2 + x} = -\infty - \sqrt{\infty - \infty} $$
To simplify, factor \( x^2 \) from inside the square root:
$$ \lim_{x \rightarrow -\infty} 2x - \sqrt{x^2\left(4 + \frac{1}{x}\right)} $$
Since \( x \to -\infty \), we have \( \sqrt{x^2} = -x \), so the expression becomes:
$$ \lim_{x \rightarrow -\infty} 2x + x \cdot \sqrt{4 + \frac{1}{x}} $$
As \( \frac{1}{x} \to 0 \), the square root tends to 2:
$$ \lim_{x \rightarrow -\infty} x(2 + 2) = \lim_{x \rightarrow -\infty} 4x = -\infty $$
Result: the limit is \( -\infty \).
Alternative explanation. For limits at infinity, the highest-degree term dominates. In the square root, the leading term is \( 4x^2 \), so: $$ \sqrt{4x^2 + x} \sim \sqrt{4x^2} = 2|x| = -2x \text{ (since } x < 0) $$ Then: $$ \lim_{x \rightarrow -\infty} 2x - (-2x) = \lim_{x \rightarrow -\infty} 4x = -\infty $$
Exercise 8
We now evaluate the following limit as \( x \to +\infty \):
$$ \lim_{x \rightarrow +\infty} \sqrt{x^2 + 3x - 4} - x $$
This is a classic example of the indeterminate form \( \infty - \infty \):
$$ \sqrt{x^2 + 3x - 4} - x = \infty - \infty $$
To eliminate this form, we multiply and divide by the conjugate:
$$ \frac{\left(\sqrt{x^2 + 3x - 4} - x\right)\left(\sqrt{x^2 + 3x - 4} + x\right)}{\sqrt{x^2 + 3x - 4} + x} $$
This simplifies to:
$$ \frac{(x^2 + 3x - 4) - x^2}{\sqrt{x^2 + 3x - 4} + x} = \frac{3x - 4}{\sqrt{x^2 + 3x - 4} + x} $$
Now we have a rational expression of type \( \infty / \infty \). We simplify by factoring:
Numerator:
$$ x(3 - \frac{4}{x}) $$
Denominator:
$$ x\left( \sqrt{1 + \frac{3}{x} - \frac{4}{x^2}} + 1 \right) $$
Cancel the common factor \( x \):
$$ \lim_{x \rightarrow +\infty} \frac{3 - \frac{4}{x}}{\sqrt{1 + \frac{3}{x} - \frac{4}{x^2}} + 1} $$
As \( x \to +\infty \), the fractions vanish, and we get:
$$ \frac{3}{2} $$
Result: the value of the limit is \( \frac{3}{2} \).
Exercise 9
We’re asked to evaluate the limit of the following rational function:
$$ \lim_{x \rightarrow -5} \frac{x^2 + 3x - 10}{x^2 - 25} $$
We begin by determining the domain:
$$ D_f = (-\infty, -5) \cup (-5, 5) \cup (5, +\infty) $$
The function is undefined at \( x = -5 \) and \( x = 5 \), where the denominator vanishes.
However, since \( x = -5 \) is a limit point of the domain, the limit can still be computed.
Substituting directly, we obtain an indeterminate form of type 0/0:
$$ \lim_{x \rightarrow -5} \frac{x^2 + 3x - 10}{x^2 - 25} = \frac{0}{0} $$
To resolve this, we factor both the numerator and denominator.
The numerator factors as:
$$ x^2 + 3x - 10 = (x - 2)(x + 5) $$
The denominator is a difference of squares:
$$ x^2 - 25 = (x - 5)(x + 5) $$
Substituting into the expression:
$$ \lim_{x \rightarrow -5} \frac{(x - 2)(x + 5)}{(x - 5)(x + 5)} $$
We cancel the common factor \( x + 5 \):
$$ \lim_{x \rightarrow -5} \frac{x - 2}{x - 5} $$
Now we can evaluate the limit directly:
$$ \frac{-5 - 2}{-5 - 5} = \frac{-7}{-10} = \frac{7}{10} $$
Final result: the limit is \( \frac{7}{10} \).
Note. Since we started with an indeterminate form, we could have also applied L’Hôpital’s Rule: $$ \lim_{x \rightarrow -5} \frac{x^2 + 3x - 10}{x^2 - 25} = \lim_{x \rightarrow -5} \frac{2x + 3}{2x} = \frac{-7}{-10} = \frac{7}{10} $$ As expected, both approaches yield the same result.
Exercise 10
We now evaluate the following limit:
$$ \lim_{x \rightarrow 0} \frac{\log \sqrt{x + 1}}{x} $$
The domain is given by:
$$ D_f = (-1, 0) \cup (0, +\infty) $$
Since \( x = 0 \) is a limit point of the domain, we can proceed.
Substituting directly, we obtain:
$$ \frac{\log \sqrt{1}}{0} = \frac{0}{0} $$
This is an indeterminate form of type 0/0.
To simplify, multiply and divide by 2:
$$ \lim_{x \rightarrow 0} \frac{2 \cdot \log \sqrt{x + 1}}{2x} $$
Using the identity \( 2 \log \sqrt{a} = \log a \), we get:
$$ \lim_{x \rightarrow 0} \frac{\log (x + 1)}{2x} = \frac{1}{2} \cdot \lim_{x \rightarrow 0} \frac{\log (x + 1)}{x} $$
It is well known that:
$$ \lim_{x \rightarrow 0} \frac{\log (1 + x)}{x} = 1 $$
Hence:
$$ \lim_{x \rightarrow 0} \frac{\log \sqrt{x + 1}}{x} = \frac{1}{2} $$
Final result: the limit is \( \frac{1}{2} \).
Note. We could also apply L’Hôpital’s Rule. The derivative of the numerator is computed via the chain rule: $$ D[\log \sqrt{x + 1}] = \frac{1}{\sqrt{x + 1}} \cdot \frac{1}{2\sqrt{x + 1}} = \frac{1}{2(x + 1)} $$ So, $$ \lim_{x \rightarrow 0} \frac{1}{2(x + 1)} = \frac{1}{2} $$
Alternative method
This limit can also be evaluated using a different strategy.
We rewrite the argument of the logarithm as \( 1 + (\sqrt{x + 1} - 1) \), which allows us to use the approximation \( \log(1 + f(x)) \approx f(x) \) for small \( f(x) \):
We then multiply and divide by \( \sqrt{x + 1} - 1 \):
$$ \lim_{x \rightarrow 0} \frac{\log(1 + (\sqrt{x + 1} - 1))}{\sqrt{x + 1} - 1} \cdot \frac{\sqrt{x + 1} - 1}{x} $$
The first factor tends to 1, by the limit \( \lim_{f \to 0} \frac{\log(1 + f)}{f} = 1 \). We are left with:
$$ \lim_{x \rightarrow 0} \frac{\sqrt{x + 1} - 1}{x} $$
This is still of the form 0/0, so we rationalize by multiplying numerator and denominator by \( \sqrt{x + 1} + 1 \):
$$ \frac{\sqrt{x + 1} - 1}{x} \cdot \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1} = \frac{(x + 1) - 1}{x(\sqrt{x + 1} + 1)} = \frac{x}{x(\sqrt{x + 1} + 1)} $$
Simplifying:
$$ \frac{1}{\sqrt{x + 1} + 1} $$
Now we substitute \( x = 0 \):
$$ \frac{1}{2} $$
Conclusion: once again, the value of the limit is \( \frac{1}{2} \).
Note. This example illustrates how a limit can be approached using various techniques. Some are faster and more elegant, others longer and more involved - yet they all lead to the same result.
Exercise 11
$$ \lim_{x \rightarrow +\infty} x + 2 - \sqrt{x^2 + 4x + 8} $$
This is an indeterminate form of type ∞ - ∞:
To resolve it, we multiply and divide by the conjugate of the expression:
$$ \lim_{x \rightarrow +\infty} (x + 2 - \sqrt{x^2 + 4x + 8}) \cdot \frac{x + 2 + \sqrt{x^2 + 4x + 8}}{x + 2 + \sqrt{x^2 + 4x + 8}} $$
This simplifies using the identity \( (a - b)(a + b) = a^2 - b^2 \):
$$ \lim_{x \rightarrow +\infty} \frac{(x + 2)^2 - (x^2 + 4x + 8)}{x + 2 + \sqrt{x^2 + 4x + 8}} $$
$$ \lim_{x \rightarrow +\infty} \frac{x^2 + 4x + 4 - x^2 - 4x - 8}{x + 2 + \sqrt{x^2 + 4x + 8}} = \lim_{x \rightarrow +\infty} \frac{-4}{x + 2 + \sqrt{x^2 + 4x + 8}} $$
The indeterminate form has been eliminated. We now compute the limit:
$$ \lim_{x \rightarrow +\infty} \frac{-4}{x + 2 + \sqrt{x^2 + 4x + 8}} = \frac{-4}{\infty} = 0 $$
Final result: the limit is 0.
Exercise 12
Evaluate the limit:
$$ \lim_{x \rightarrow -2} \frac{3x^2 + x - 10}{x^2 - 5x - 14} $$
To find the domain, we determine where the denominator is zero. Solving:
$$ x = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} = \frac{5 \pm 9}{2} $$
So the excluded values are:
$$ x = -2 \quad \text{and} \quad x = 7 $$
Thus, the domain is:
$$ D_f = (-\infty, -2) \cup (-2, 7) \cup (7, +\infty) $$
Since \( x = -2 \) is a limit point, the limit can be evaluated.
Substituting directly gives a 0/0 indeterminate form.
We factor both numerator and denominator using synthetic division:
Numerator: \( (x + 2)(3x - 5) \) Denominator: \( (x + 2)(x - 7) \)
Substituting:
$$ \lim_{x \rightarrow -2} \frac{(x + 2)(3x - 5)}{(x + 2)(x - 7)} $$
Canceling the common factor:
$$ \lim_{x \rightarrow -2} \frac{3x - 5}{x - 7} = \frac{-11}{-9} = \frac{11}{9} $$
Final result: the limit is \( \frac{11}{9} \).
Note. This limit can also be solved using L’Hôpital’s Rule: $$ \lim_{x \rightarrow -2} \frac{3x^2 + x - 10}{x^2 - 5x - 14} = \lim_{x \rightarrow -2} \frac{6x + 1}{2x - 5} = \frac{-11}{-9} = \frac{11}{9} $$ Same result, confirming the validity of the simplification.
Exercise 13
Let’s evaluate the limit:
$$ \lim_{x \rightarrow 2} \frac{\sqrt{x^2 - 2x + 9} - 3}{x^3 - x^2 - x - 2} $$
We begin by factoring the denominator via synthetic division:
$$ x^3 - x^2 - x - 2 = (x - 2)(x^2 + x + 1) $$
So the expression becomes:
$$ \lim_{x \rightarrow 2} \frac{\sqrt{x^2 - 2x + 9} - 3}{(x - 2)(x^2 + x + 1)} $$
Substituting gives a 0/0 indeterminate form.
To resolve it, we multiply by the conjugate of the numerator:
$$ \cdot \frac{\sqrt{x^2 - 2x + 9} + 3}{\sqrt{x^2 - 2x + 9} + 3} $$
The numerator becomes a difference of squares:
$$ (x^2 - 2x + 9) - 9 = x^2 - 2x $$
Factoring:
$$ \frac{x(x - 2)}{(x - 2)(x^2 + x + 1)(\sqrt{x^2 - 2x + 9} + 3)} $$
Canceling the common factor \( (x - 2) \):
$$ \lim_{x \rightarrow 2} \frac{x}{(x^2 + x + 1)(\sqrt{x^2 - 2x + 9} + 3)} $$
Now substitute \( x = 2 \):
$$ \frac{2}{(4 + 2 + 1)(\sqrt{9} + 3)} = \frac{2}{7 \cdot 6} = \frac{1}{21} $$
Final result: the limit is \( \frac{1}{21} \).
Exercise 14
We are asked to evaluate the limit:
$$ \lim_{x \rightarrow -1} \frac{\sqrt{x^2 + 3} - 2}{3 - \sqrt{8 - x^3}} $$
The function is defined on the interval:
$$ D_f = (-\infty , -1) \cup (-1, 2) $$
Since \( x = -1 \) is a limit point of the domain, we can attempt to compute the limit.
Substituting directly, we get the indeterminate form 0/0:
$$ \frac{\sqrt{1 + 3} - 2}{3 - \sqrt{8 - (-1)^3}} = \frac{2 - 2}{3 - \sqrt{9}} = \frac{0}{0} $$
To resolve the indeterminacy, we begin by rationalizing the numerator:
$$ \lim_{x \rightarrow -1} \frac{\sqrt{x^2 + 3} - 2}{3 - \sqrt{8 - x^3}} \cdot \frac{\sqrt{x^2 + 3} + 2}{\sqrt{x^2 + 3} + 2} $$
Using the identity \( (a - b)(a + b) = a^2 - b^2 \), we simplify:
$$ \lim_{x \rightarrow -1} \frac{(x^2 + 3) - 4}{(3 - \sqrt{8 - x^3})(\sqrt{x^2 + 3} + 2)} = \lim_{x \rightarrow -1} \frac{x^2 - 1}{(3 - \sqrt{8 - x^3})(\sqrt{x^2 + 3} + 2)} $$
We now split the expression into two separate limits:
$$ \left( \lim_{x \rightarrow -1} \frac{x^2 - 1}{3 - \sqrt{8 - x^3}} \right) \cdot \left( \lim_{x \rightarrow -1} \frac{1}{\sqrt{x^2 + 3} + 2} \right) $$
The second factor evaluates easily to \( \frac{1}{4} \). We now turn our attention to the first limit:
$$ \frac{1}{4} \cdot \lim_{x \rightarrow -1} \frac{x^2 - 1}{3 - \sqrt{8 - x^3}} $$
To eliminate the remaining indeterminacy, we rationalize the denominator:
$$ \frac{1}{4} \cdot \lim_{x \rightarrow -1} \frac{x^2 - 1}{3 - \sqrt{8 - x^3}} \cdot \frac{3 + \sqrt{8 - x^3}}{3 + \sqrt{8 - x^3}} $$
Simplifying the denominator using the difference of squares:
$$ \frac{1}{4} \cdot \lim_{x \rightarrow -1} \frac{(x^2 - 1)(3 + \sqrt{8 - x^3})}{9 - (8 - x^3)} = \frac{1}{4} \cdot \lim_{x \rightarrow -1} \frac{(x^2 - 1)(3 + \sqrt{8 - x^3})}{1 + x^3} $$
We now recognize that \( x^2 - 1 = (x - 1)(x + 1) \) and \( 1 + x^3 = (x + 1)(1 - x + x^2) \), so:
$$ \frac{1}{4} \cdot \lim_{x \rightarrow -1} \frac{(x - 1)(x + 1)(3 + \sqrt{8 - x^3})}{(x + 1)(1 - x + x^2)} $$
Canceling the common factor \( x + 1 \):
$$ \frac{1}{4} \cdot \lim_{x \rightarrow -1} \frac{(x - 1)(3 + \sqrt{8 - x^3})}{1 - x + x^2} $$
Now we substitute \( x = -1 \):
Numerator: \( (-2)(3 + \sqrt{9}) = (-2)(6) = -12 \)
Denominator: \( 1 - (-1) + 1 = 3 \)
Final value:
$$ \frac{1}{4} \cdot \frac{-12}{3} = -1 $$
Conclusion: the limit evaluates to \( -1 \).
Exercise 15
We are asked to compute the limit:
$$ \lim_{x \rightarrow 0} \frac{\sin(x)}{x} $$
This is a classic special limit, and one of the most fundamental in analysis.
We recognize immediately that it yields an indeterminate form 0/0.
To justify the result analytically, we apply the first-order Maclaurin expansion of \( \sin(x) \):
$$ \sin(x) = x + o(x) $$
Substituting into the limit:
$$ \lim_{x \rightarrow 0} \frac{x + o(x)}{x} $$
Factoring \( x \) in the numerator:
$$ \lim_{x \rightarrow 0} \frac{x(1 + o(1))}{x} = \lim_{x \rightarrow 0} (1 + o(1)) $$
Since \( o(1) \to 0 \) as \( x \to 0 \), we conclude:
$$ \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 $$
Conclusion: the limit is equal to \( 1 \).
Exercise 16
Let’s evaluate the limit:
$$ \lim_{x \to 0} \frac{\sin(3x)}{x} $$
This expression closely resembles the standard identity:
$$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 $$
However, it’s important to notice that the argument of the sine function does not match the denominator.
The key is to express the function in a form where we can apply the identity. To do that, we multiply and divide by 3:
$$ \lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot 3 $$
Now we clearly have the standard form \( \frac{\sin(kx)}{kx} \to 1 \) as \( x \to 0 \), with \( k = 3 \):
$$ \lim_{x \to 0} \frac{\sin(3x)}{3x} = 1 $$
So the final result is:
$$ 3 \cdot 1 = 3 $$
Conclusion: the limit is \( 3 \).
This example reinforces the idea that recognizing the structure of a known limit - especially when disguised - is often more important than simply memorizing the result. The trick is to manipulate the expression so the argument of the sine matches the denominator.
Exercise 17
We’re asked to evaluate the limit:
\[ \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} \]
This is an indeterminate form of type \( \frac{0}{0} \), since both the numerator and denominator tend to zero as \( x \to 0 \).
To resolve the indeterminacy, we multiply by the conjugate of the numerator to eliminate the square root:
\[ \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1} \]
Applying the identity \( (a - b)(a + b) = a^2 - b^2 \), we obtain:
\[ \lim_{x \to 0} \frac{(1 + x) - 1}{x(\sqrt{1 + x} + 1)} \]
\[ \lim_{x \to 0} \frac{x}{x(\sqrt{1 + x} + 1)} \]
Canceling the common factor of \( x \):
\[ \lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} \]
Now the expression is continuous at \( x = 0 \), so we substitute directly:
\[ \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2} \]
Final result: the limit evaluates to \( \frac{1}{2} \).
Exercise 18
We now evaluate the limit:
\[ \lim_{x \to 0} \frac{1 - \cos(3x)}{x^2} \]
This expression resembles the standard identity:
\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \]
To bring it into this form, we perform a change of variable: let \( u = 3x \).
Then as \( x \to 0 \), we also have \( u \to 0 \), and the limit becomes:
\[ \lim_{u \to 0} \frac{1 - \cos(u)}{x^2} \]
Since \( x = \frac{u}{3} \), we substitute accordingly:
\[ \lim_{u \to 0} \frac{1 - \cos(u)}{ \left( \frac{u}{3} \right)^2 } = \lim_{u \to 0} \frac{1 - \cos(u)}{ \frac{u^2}{9} } \]
We now simplify the expression:
\[ \lim_{u \to 0} \frac{1 - \cos(u)}{u^2} \cdot 9 \]
Applying the standard limit:
\[ \lim_{u \to 0} \frac{1 - \cos(u)}{u^2} = \frac{1}{2} \Rightarrow 9 \cdot \frac{1}{2} = \frac{9}{2} \]
Final result: the value of the limit is \( \frac{9}{2} \).
Exercise 19
We are asked to compute the limit:
$$ \lim_{x \to 0} \frac{\sin(x) - x}{x^3} $$
This too is an indeterminate form of type 0/0.
To resolve it, we use the third-order Maclaurin expansion of \( \sin(x) \), which is suitable because the denominator is a cubic polynomial:
$$ \sin(x) = x - \frac{x^3}{3!} + o(x^3) = x - \frac{x^3}{6} + o(x^3) $$
Note. We truncate the series at degree three because the denominator is \( x^3 \). Higher-order terms are negligible in this context.
Substitute into the limit expression:
$$ \lim_{x \to 0} \frac{\left( x - \frac{x^3}{6} + o(x^3) \right) - x}{x^3} $$
$$ \lim_{x \to 0} \frac{-\frac{x^3}{6} + o(x^3)}{x^3} $$
We now factor out \( x^3 \) from the numerator:
$$ \lim_{x \to 0} \frac{x^3 \left( -\frac{1}{6} + o(1) \right)}{x^3} $$
Canceling \( x^3 \):
$$ \lim_{x \to 0} -\frac{1}{6} + o(1) $$
Since \( o(1) \to 0 \) as \( x \to 0 \), we conclude:
$$ \lim_{x \to 0} \frac{\sin(x) - x}{x^3} = -\frac{1}{6} $$
Final result: the limit is \( -\frac{1}{6} \).
Exercise 20
We are asked to evaluate the following limit:
$$ \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^2} $$
This is an indeterminate form of type 0/0:
$$ \frac{1 - \cos(0)}{0^2} = \frac{0}{0} $$
There are several effective techniques for resolving this limit.
Solution using the Maclaurin series
Since the limit is taken as \( x \to 0 \), we can replace the cosine function with its second-order Maclaurin expansion:
$$ \cos(x) = 1 - \frac{x^2}{2} + o[x^2] $$
Substituting into the original expression yields:
$$ \lim_{x \rightarrow 0} \frac{1 - (1 - \frac{x^2}{2} + o[x^2])}{x^2} $$
$$ \lim_{x \rightarrow 0} \frac{\frac{x^2}{2} - o[x^2]}{x^2} $$
$$ \lim_{x \rightarrow 0} \left( \frac{1}{2} - o[1] \right) = \frac{1}{2} $$
Conclusion: the value of the limit is \( \frac{1}{2} \).
Solution using L’Hôpital’s Rule
We can also apply L’Hôpital’s Rule, which is valid in this indeterminate case:
First derivative:
$$ \lim_{x \rightarrow 0} \frac{\sin(x)}{2x} = \frac{0}{0} $$
Second derivative:
$$ \lim_{x \rightarrow 0} \frac{\cos(x)}{2} = \frac{1}{2} $$
Conclusion: applying the rule twice confirms that the limit is \( \frac{1}{2} \).
Solution using trigonometric identities
We use the double-angle identity for cosine:
$$ \cos(x) = 1 - 2\sin^2\left( \frac{x}{2} \right) $$
Substituting into the original limit gives:
$$ \lim_{x \rightarrow 0} \frac{2\sin^2\left( \frac{x}{2} \right)}{x^2} $$
Factor out the constant:
$$ 2 \cdot \lim_{x \rightarrow 0} \left( \frac{\sin\left( \frac{x}{2} \right)}{x} \right)^2 $$
We rewrite the denominator to match the argument of the sine function:
$$ 2 \cdot \lim_{x \rightarrow 0} \left( \frac{1}{2} \cdot \frac{\sin\left( \frac{x}{2} \right)}{\frac{x}{2}} \right)^2 $$
$$ 2 \cdot \left( \frac{1}{2} \right)^2 \cdot \lim_{x \rightarrow 0} \left( \frac{\sin\left( \frac{x}{2} \right)}{\frac{x}{2}} \right)^2 = \frac{1}{2} \cdot 1^2 = \frac{1}{2} $$
Conclusion: the limit evaluates to \( \frac{1}{2} \), consistent with all previous methods.
Exercise 21
We now evaluate the limit:
$$ \lim_{x \rightarrow 0} \frac{\cos(x) - 1}{x} $$
Again, we are dealing with an indeterminate form of type 0/0.
Solution using the Maclaurin series
We expand \( \cos(x) \) using its second-order Maclaurin polynomial:
$$ \cos(x) = 1 - \frac{x^2}{2} + o[x^2] $$
Substituting into the limit:
$$ \lim_{x \rightarrow 0} \frac{(1 - \frac{x^2}{2} + o[x^2]) - 1}{x} $$
$$ \lim_{x \rightarrow 0} \frac{-\frac{x^2}{2} + o[x^2]}{x} $$
$$ \lim_{x \rightarrow 0} -\frac{x}{2} + x \cdot o[1] = 0 $$
Conclusion: the limit is 0.
Note. A first-order expansion, \( \cos(x) = 1 + o[x] \), also leads to the same result with less effort.
Solution using L’Hôpital’s Rule
Applying the rule directly:
$$ \lim_{x \rightarrow 0} \frac{D[\cos(x) - 1]}{D[x]} = \lim_{x \rightarrow 0} \frac{-\sin(x)}{1} = 0 $$
Conclusion: the limit evaluates to 0.
Solution via the definition of derivative
This expression matches the definition of the derivative of \( \cos(x) \) at \( x = 0 \):
$$ \lim_{x \rightarrow 0} \frac{\cos(x) - \cos(0)}{x - 0} = \cos'(0) = -\sin(0) = 0 $$
Conclusion: the limit is 0.
Solution using standard limits
We manipulate the expression algebraically using a standard limit:
$$ \lim_{x \rightarrow 0} \frac{\cos(x) - 1}{x} = \lim_{x \rightarrow 0} \left( -\frac{1 - \cos(x)}{x^2} \cdot x \right) $$
Since \( \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \), we get:
$$ -\frac{1}{2} \cdot 0 = 0 $$
Conclusion: the limit is 0.
Exercise 22
We are asked to compute the following limit:
$$ \lim_{x \rightarrow \frac{\pi}{4}} \frac{\tan(x)-1}{x - \frac{\pi}{4}} $$
Since the tangent of \( \pi/4 \) is:
$$ \tan\left(\frac{\pi}{4}\right) = 1 $$
we are dealing with an indeterminate form of type 0/0:
$$ \frac{\tan(x) - 1}{x - \frac{\pi}{4}} = \frac{0}{0} $$
This is the definition of the derivative of \( f(x) = \tan(x) \) evaluated at \( x = \frac{\pi}{4} \):
$$ \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} = f'(a) $$
So:
$$ \lim_{x \rightarrow \frac{\pi}{4}} \frac{\tan(x) - \tan\left(\frac{\pi}{4}\right)}{x - \frac{\pi}{4}} = \frac{d}{dx}\left[\tan(x)\right]\Big|_{x = \frac{\pi}{4}} $$
The derivative of the tangent function is:
$$ \frac{d}{dx} \tan(x) = \frac{1}{\cos^2(x)} $$
Evaluating at \( x = \frac{\pi}{4} \):
$$ \frac{1}{\left(\frac{\sqrt{2}}{2}\right)^2} = \frac{1}{\frac{1}{2}} = 2 $$
Conclusion: the limit evaluates to 2.
Alternative approach. This limit can also be solved using L’Hôpital’s Rule: $$ \lim_{x \rightarrow \frac{\pi}{4}} \frac{\tan(x) - 1}{x - \frac{\pi}{4}} = \frac{d}{dx}[\tan(x)] \bigg/ \frac{d}{dx}[x] = \frac{1}{\cos^2(x)} $$ At \( x = \frac{\pi}{4} \), this becomes: $$ \frac{1}{\left(\frac{\sqrt{2}}{2}\right)^2} = 2 $$ The result is consistent with the previous method.
Exercise 23
We now consider the limit:
$$ \lim_{x \rightarrow 0} \frac{\sin x}{e^x - 1} $$
This expression also results in an indeterminate form of type 0/0.
To evaluate it, we approximate the transcendental functions using the Maclaurin series near zero:
$$ \sin(x) = x + o[x], \quad e^x = 1 + x + o[x] $$
Substituting into the limit:
$$ \frac{x + o[x]}{(1 + x + o[x]) - 1} = \frac{x + o[x]}{x + o[x]} $$
Noting that \( o[x] = x \cdot o[1] \), we factor out \( x \):
$$ \frac{x(1 + o[1])}{x(1 + o[1])} = \frac{1 + o[1]}{1 + o[1]} $$
As \( o[1] \to 0 \), the limit becomes:
$$ \lim_{x \to 0} \frac{1 + o[1]}{1 + o[1]} = 1 $$
Conclusion: the value of the limit is 1.
Exercise 24
Let’s evaluate the following limit as \( x \to 2 \):
$$ \lim_{x \rightarrow 2} \frac{\sqrt{x^2 + 5} - \sqrt{x^3 + 1}}{2x - x^2} $$
This also leads to an indeterminate form 0/0:
$$ \frac{\sqrt{9} - \sqrt{9}}{4 - 4} = \frac{0}{0} $$
To resolve it, we rationalize the numerator:
$$ \cdot \frac{\sqrt{x^2 + 5} + \sqrt{x^3 + 1}}{\sqrt{x^2 + 5} + \sqrt{x^3 + 1}} $$
This gives:
$$ \frac{(x^2 + 5) - (x^3 + 1)}{2x - x^2} \cdot \frac{1}{\sqrt{x^2 + 5} + \sqrt{x^3 + 1}} $$
Which simplifies to:
$$ \frac{x^2 - x^3 + 4}{2x - x^2} \cdot \frac{1}{\sqrt{x^2 + 5} + \sqrt{x^3 + 1}} $$
The second factor tends to \( \frac{1}{6} \), so we write:
$$ \frac{1}{6} \cdot \lim_{x \rightarrow 2} \frac{x^2 - x^3 + 4}{2x - x^2} $$
This limit is still of type 0/0, so we factor the numerator using Ruffini’s method:
$$ \frac{1}{6} \cdot \frac{(x - 2)(-x^2 - x - 2)}{2x - x^2} $$
Note. Using Ruffini’s rule, the polynomial \( x^2 - x^3 + 4 \) can be factored as \( (x - 2)(-x^2 - x - 2) \): $$ \begin{array}{c|rrrr} & -1 & 1 & 0 & 4 \\ 2 & & -2 & -2 & -4 \\ \hline & -1 & -1 & -2 & 0 \end{array} $$
The denominator \( 2x - x^2 \) can be rewritten as \( -x(x - 2) \), allowing for simplification:
$$ \frac{1}{6} \cdot \frac{(x - 2)(-x^2 - x - 2)}{-x(x - 2)} $$
Canceling \( x - 2 \):
$$ \frac{1}{6} \cdot \frac{-x^2 - x - 2}{-x} $$
Evaluating at \( x = 2 \):
$$ \frac{1}{6} \cdot \frac{-4 - 2 - 2}{-2} = \frac{1}{6} \cdot 4 = \frac{2}{3} $$
Conclusion: the value of the limit is \( \frac{2}{3} \).
Exercise 25
In this exercise, we aim to compute the limit of the following expression as \( x \to 0 \):
$$ \lim_{x \rightarrow 0 } \frac{3 - \sqrt{9 - x^2}}{2 \sqrt{1 + x^2} - \sqrt{4 + x^2}} $$
The point \( x = 0 \) lies within the domain of the function and is a limit point.
Substituting directly leads to an indeterminate form of type 0/0:
$$ \frac{3 - \sqrt{9 - x^2}}{2 \sqrt{1 + x^2} - \sqrt{4 + x^2}} = \frac{0}{0} $$
To resolve this, we rationalize the denominator by multiplying by its conjugate:
$$ \cdot \frac{2 \sqrt{1 + x^2} + \sqrt{4 + x^2}}{2 \sqrt{1 + x^2} + \sqrt{4 + x^2}} $$
Applying the identity for the difference of squares:
$$ \frac{(3 - \sqrt{9 - x^2})(2 \sqrt{1 + x^2} + \sqrt{4 + x^2})}{4(1 + x^2) - (4 + x^2)} $$
$$ = \frac{(3 - \sqrt{9 - x^2})(2 \sqrt{1 + x^2} + \sqrt{4 + x^2})}{3x^2} $$
We split the expression as a product of two limits:
$$ \left( \frac{3 - \sqrt{9 - x^2}}{3x^2} \right) \cdot \left( 2 \sqrt{1 + x^2} + \sqrt{4 + x^2} \right) $$
As \( x \to 0 \), the second factor approaches 4. So we get:
$$ 4 \cdot \lim_{x \to 0} \frac{3 - \sqrt{9 - x^2}}{3x^2} $$
We now rationalize the numerator of the remaining limit:
$$ \cdot \frac{3 + \sqrt{9 - x^2}}{3 + \sqrt{9 - x^2}} $$
Using the difference of squares again:
$$ \frac{9 - (9 - x^2)}{3x^2(3 + \sqrt{9 - x^2})} = \frac{x^2}{3x^2(3 + \sqrt{9 - x^2})} $$
Simplifying:
$$ \frac{1}{3(3 + \sqrt{9 - x^2})} $$
Taking the limit as \( x \to 0 \):
$$ 4 \cdot \frac{1}{3(3 + 3)} = 4 \cdot \frac{1}{18} = \frac{2}{9} $$
Conclusion: the limit evaluates to \( \frac{2}{9} \).
Exercise 26
Let’s compute the following limit as \( x \to 0 \):
$$ \lim_{x \rightarrow 0} \frac{(e^{3x} - 1) \sin(x^4)}{x^2 \log(1 + x^3)} $$
Substituting directly gives the indeterminate form 0/0.
We begin by rewriting the expression as a product of two simpler limits:
$$ \left( \frac{e^{3x} - 1}{x} \right) \cdot \left( \frac{\sin(x^4)}{x \log(1 + x^3)} \right) $$
We introduce a factor of 3 to isolate a standard limit:
$$ 3 \cdot \left( \frac{e^{3x} - 1}{3x} \right) \cdot \left( \frac{\sin(x^4)}{x \log(1 + x^3)} \right) $$
We now recognize:
\( \lim_{t \to 0} \frac{e^t - 1}{t} = 1 \) with \( t = 3x \)
So we are left with:
$$ 3 \cdot 1 \cdot \lim_{x \to 0} \frac{\sin(x^4)}{x \log(1 + x^3)} $$
We factor the denominator further by multiplying and dividing by \( x^3 \):
$$ = 3 \cdot \frac{\sin(x^4)}{x^4} \cdot \frac{1}{\frac{\log(1 + x^3)}{x^3}} $$
Now we apply two well-known limits:
- \( \lim_{t \to 0} \frac{\sin(t)}{t} = 1 \), with \( t = x^4 \)
- \( \lim_{t \to 0} \frac{\log(1 + t)}{t} = 1 \), with \( t = x^3 \)
Therefore:
$$ 3 \cdot 1 \cdot \frac{1}{1} = 3 $$
Conclusion: the limit is equal to 3.
Exercise 27
We are asked to evaluate the limit of the function:
$$ \lim_{x \to \infty} \frac{x + \sin(x)}{x - \cos(x)} $$
Although this appears to be of type \( \frac{\infty}{\infty} \), L’Hôpital’s Rule cannot be applied, as the derivative of the numerator and denominator does not yield a determinate form:
$$ \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \frac{1 + \cos(x)}{1 + \sin(x)} $$ which does not converge due to the oscillatory behavior of sine and cosine.
Method 1: Algebraic simplification
Divide numerator and denominator by \( x \):
$$ \frac{x + \sin(x)}{x - \cos(x)} = \frac{1 + \frac{\sin(x)}{x}}{1 - \frac{\cos(x)}{x}} $$
As \( x \to \infty \), \( \frac{\sin(x)}{x} \to 0 \) and \( \frac{\cos(x)}{x} \to 0 \), since sine and cosine remain bounded.
Hence:
$$ \lim_{x \to \infty} \frac{1 + \frac{\sin(x)}{x}}{1 - \frac{\cos(x)}{x}} = \frac{1 + 0}{1 - 0} = 1 $$
Conclusion: the limit is 1.
Method 2: Using the squeeze theorem
Let:
$$ f(x) = \frac{x + \sin(x)}{x - \cos(x)} $$
Since \( -1 \leq \sin(x) \leq 1 \) and \( -1 \leq \cos(x) \leq 1 \), we have:
$$ \frac{x - 1}{x + 1} \leq f(x) \leq \frac{x + 1}{x - 1} $$
Taking limits of the bounds as \( x \to \infty \):
$$ \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}} \leq f(x) \leq \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} $$
Both sides converge to 1, so by the squeeze theorem:
$$ \lim_{x \to \infty} \frac{x + \sin(x)}{x - \cos(x)} = 1 $$
