Limits of Powers
If a function has a finite limit, $$ \lim_{x \to a} f(x) = l $$ then any positive integer power of the function also has a limit, given by the corresponding power of the limit: $$ \lim_{x \to a} [f(x)]^n = \left( \lim_{x \to a} f(x) \right)^n = l^n $$ where the exponent $ n $ is a nonzero natural number, $ n \in \mathbb{N} - {0} $.
Equivalently, the limit of a power equals the power of the limit.
This statement is a direct consequence of the product limit theorem.
The intuition is immediate. When a function converges to a value, its integer powers converge to the corresponding powers of that value.
A Practical Example
Consider the function
$$ f(x) = 2x - 1 $$
We compute the limit
$$ \lim_{x \to 3} [f(x)]^4 $$
First evaluate the limit of the function itself:
$$ \lim_{x \to 3} (2x - 1) $$
The function is continuous at $ x = 3 $, so direct substitution is valid: $ 2 \cdot 3 - 1 = 6 - 1 = 5 $. Hence:
$$ \lim_{x \to 3} (2x - 1) = 5 $$
Apply the power rule:
$$ \lim_{x \to 3} [f(x)]^4 = \left( \lim_{x \to 3} (2x - 1) \right)^4 = 5^4 = 625 $$
The result follows with minimal computation.
Example 2
Consider the function
$$ f(x)=\frac{x+1}{x-2} $$
We compute the limit
$$ \lim_{x \to 4} \left(\frac{x+1}{x-2}\right)^3 $$
Begin with the limit of the function. Since the function is continuous at $ x = 4 $, substitute directly:
$ \frac{4+1}{4-2} = \frac{5}{2} $. Therefore:
$$ \lim_{x \to 4} \frac{x+1}{x-2} = \frac{5}{2} $$
Now apply the power rule:
$$ \lim_{x \to 4} \left(\frac{x+1}{x-2}\right)^3 = \left( \lim_{x \to 4} \frac{x+1}{x-2} \right)^3 = \left(\frac{5}{2}\right)^3=\frac{125}{8} $$
Hence:
$$ \lim_{x \to 4} \left(\frac{x+1}{x-2}\right)^3=\frac{125}{8} $$
Once again, the theorem reduces the calculation to a few straightforward steps.
Proof
The argument relies on a basic structural fact.
The n-th power of a function is the product of the function by itself n times:
$$ [f(x)]^n = f(x) \cdot f(x) \cdot \dots \cdot f(x) $$
If the limit of $ f(x) $ exists, the product limit theorem applies:
$$ \lim (f(x)\cdot g(x)) = \lim f(x) \cdot \lim g(x) $$
Iterating this property n times gives:
$$ \lim [f(x)]^n = (\lim f(x))^n $$
The essential observation is that exponentiation by a natural number corresponds to repeated multiplication.
Does the Power Rule Hold for Infinite Limits?
When the limit is $ +\infty $ or $ -\infty $, the rule remains valid, but the outcome depends on the sign of the limit and the parity of the exponent.
- If $$ \lim_{x \to a} f(x) = +\infty $$ then $$ \lim_{x \to a} [f(x)]^n = (+\infty)^n $$ An unbounded positive quantity remains positive and diverges when raised to any positive natural power. Therefore, $ (+\infty)^n = +\infty $ for every natural $ n $.
Example: $$ \lim_{x \to +\infty} x^5 $$ Since $$ \lim_{x \to +\infty} x = +\infty $$ it follows that $$ \lim_{x \to +\infty} x^5 = +\infty $$
- If $$ \lim_{x \to a} f(x) = -\infty $$ then
$$ \lim_{x \to a} [f(x)]^n = \begin{cases} +\infty & \text{if } n \text{ is even} \\ \\ -\infty & \text{if } n \text{ is odd} \end{cases} $$
An even exponent yields a positive result, whereas an odd exponent preserves the negative sign.
Example: $$ \lim_{x \to -\infty} x^2 $$ Although $$ \lim_{x \to -\infty} x = -\infty $$ the square is positive, hence $$ \lim_{x \to -\infty} x^2 = +\infty $$ With an odd exponent, $$ \lim_{x \to -\infty} x^3 = -\infty $$
What If the Exponent Is a Function?
The power rule applies directly when the exponent is constant. If the exponent depends on $ x $, the rule cannot be applied without further analysis.
In such cases, evaluating the limit is justified only when no indeterminate form appears, such as $ 0^0 $, $ 1^\infty $, or $ \infty^0 $.
Whenever an indeterminate form arises, direct substitution is inadequate. Additional analytical tools are required.
Example
Consider the limit
$$ \lim_{x \to 2} (x+1)^{3x} $$
Evaluate the limits separately:
$$ \lim_{x \to 2} (x+1) = 3 $$
$$ \lim_{x \to 2} (3x) = 6 $$
The expression becomes $ 3^6 $, which is determinate. Therefore:
$$ \lim (x+1)^{3x} = 3^6 = 729 $$
The rule applies because the resulting form is well defined.
Example 2
Consider the limit
$$ \lim_{x \to +\infty} \left(1+\frac{1}{x}\right)^x $$
Compute the limits separately:
$$ \lim_{x \to +\infty} \left(1+\frac{1}{x}\right) = 1 $$
$$ \lim_{x \to +\infty} x = \infty $$
This produces the indeterminate form $ 1^\infty $:
$$ \lim_{x \to +\infty} \left(1+\frac{1}{x}\right)^x = 1^\infty $$
Therefore, the power rule cannot be invoked mechanically.
Note. A superficial interpretation would suggest $ 1^\infty = 1 $, which is incorrect. The correct value is $$ \lim_{x \to +\infty} \left(1+\frac{1}{x}\right)^x = e $$ This is a remarkable limit, established through a specific proof.
And so on.
