Limit of the Product of Two Functions

Let $ f(x) $ and $ g(x) $ be functions that admit finite limits at the same point \( x_0 \): \[ \lim_{x \to x_0} f(x) = l \in \mathbb{R} \] \[ \lim_{x \to x_0} g(x) = m \in \mathbb{R} \] Then the limit of their product equals the product of the limits: \[ \lim_{x \to x_0} [f(x)\cdot g(x)] = l \cdot m \]

This statement is a cornerstone of limit theory.

As \( x \) approaches \( x_0 \), \( f(x) \) approaches \( l \), and \( g(x) \) approaches \( m \).

\[ f(x) \approx l \quad \text{and} \quad g(x) \approx m \;\Longrightarrow\; f(x)g(x) \approx l m. \]

Accordingly, the product \( f(x)g(x) \) approaches \( l \cdot m \).

The Case of a Constant Times a Function

If $ g(x)=k $ is constant, we obtain an important operational rule:

$$ \lim_{x \to x_0} [k \cdot f(x)] = k \cdot \lim_{x \to x_0} f(x) = k \cdot l $$

This follows because multiplying by a constant scales every value of the function by that constant.

Therefore, if \( f(x) \to l \) as \( x \to x_0 \), then \( k f(x) \to k l \).

A Practical Example

Consider the functions

$$ f(x) = x $$

$$ g(x) = x+1 $$

Evaluate the limits as \( x \to 2 \):

\[ \lim_{x \to 2} f(x) = 2 \]

\[ \lim_{x \to 2} g(x) = 3 \]

Now compute the limit of the product:

\[ \lim_{x \to 2} f(x) \cdot g(x)  \]

\[ \lim_{x \to 2} x \cdot (x+1) = 2 \cdot (2+1) = 6  \]

Applying the product rule leads to the same result:

\[ \lim_{x \to 2} [x(x+1)] = 2 \cdot 3 = 6 \]

The agreement between the two methods confirms the theorem.

Proof

The argument is developed in two steps. We begin with the simplest case \( l = m = 0 \), then proceed to the general case.

1] Simple Case: \( l = m = 0 \)

Consider two functions whose limits vanish as $ x \to x_0 $:

\[ \lim_{x \to x_0} f(x) = 0 \]

\[ \lim_{x \to x_0} g(x) = 0 \]

By the definition of limit, for every arbitrarily small positive number \( \varepsilon > 0 \), there exists a neighborhood of \( x_0 \) such that:

\[ |f(x)| < \varepsilon \quad \text{and} \quad |g(x)| < \varepsilon \]

If both inequalities hold within the same neighborhood, we may multiply them:

\[ |f(x)g(x)| = |f(x)| \, |g(x)| < \varepsilon \cdot \varepsilon = \varepsilon^2 \]

Because \( \varepsilon \) is arbitrary, \( \varepsilon^2 \) can be made arbitrarily small. Hence:

\[ \lim_{x \to x_0} f(x)g(x) = 0 \]

This proves that when both limits are zero, the limit of the product is also zero.

2] General Case: arbitrary \( l \) and \( m \)

Assume now that both limits are finite:

\[ \lim_{x \to x_0} f(x) = l \]

\[ \lim_{x \to x_0} g(x) = m \]

Shift each function so that the new limits are zero:

\[ \lim_{x \to x_0} (f(x) - l) = 0 \]

\[ \lim_{x \to x_0} (g(x) - m) = 0 \]

Therefore, the limit of their product is zero:

\[ \lim_{x \to x_0} (f(x) - l)(g(x) - m) = 0 \]

Expand the product:

\[ (f(x) - l)(g(x) - m) =  f(x)g(x) - m f(x) - l g(x) + lm \]

Rearrange to isolate \( f(x)g(x) \):

\[ f(x)g(x) = (f(x) - l)(g(x) - m)  + mf(x) + lg(x) - lm \]

Take the limit as $ x \to x_0 $ on both sides:

\[ \lim_{x \to x_0} f(x)g(x) = \lim_{x \to x_0} \left[(f(x) - l)(g(x) - m)  + mf(x) + lg(x) - lm \right] \]

Apply the sum rule for limits:

\[ \lim_{x \to x_0} f(x)g(x) = \lim_{x \to x_0} (f(x) - l)(g(x) - m)  + \lim_{x \to x_0} mf(x) + \lim_{x \to x_0} lg(x) - \lim_{x \to x_0} lm \]

Since \( \lim_{x \to x_0}(f(x) - l)(g(x) - m) = 0 \):

\[ \lim_{x \to x_0} f(x)g(x) = 0  + \lim_{x \to x_0} mf(x) + \lim_{x \to x_0} lg(x) - \lim_{x \to x_0} lm \]

Using the constant multiple rule:

\[ \lim_{x \to x_0} mf(x) = m \lim_{x \to x_0} f(x) = ml = lm \]

\[ \lim_{x \to x_0} lg(x) = l \lim_{x \to x_0} g(x) = lm \]

Moreover, the limit of a constant equals the constant itself:

\[ \lim_{x \to x_0} lm = lm \]

Substituting these results:

\[ \lim_{x \to x_0} f(x)g(x) = 0  + lm + lm - lm \]

Therefore:

\[ \lim_{x \to x_0} f(x)g(x) = lm \]

As required 

Product of a Constant and a Function

Let \( k \in \mathbb{R} \) with \( k \neq 0 \). If the function $ f(x) $ admits the finite limit \[ \lim_{x \to a} f(x) = l \] then the limit of the product equals the constant times the limit of the function: \[ \lim_{x \to a} [k \cdot f(x)] = k \cdot \lim_{x \to a} f(x) = k l \]

In essence, multiplying a function by a constant scales each of its values by that constant.

Therefore, if \( f(x) \to l \) as \( x \to x_0 \), then \( k f(x) \to k l \).

The constant does not influence the limiting process itself, only the resulting value.

Practical example

Consider the function

\[ f(x) = x^2 - 1 \]

Compute the limit as \( x \to 2 \).

The function is continuous at \( x_0 = 2 \), so direct substitution is valid:

\[ \lim_{x \to 2} (x^2 - 1) = 2^2 - 1 = 4 - 1 = 3 \]

Hence,

\[ \lim_{x \to 2} f(x) = 3 \]

Now multiply the function by a positive constant:

\[ 5 f(x) = 5(x^2 - 1) \]

Evaluate the limit:

\[ \lim_{x \to 2} 5(x^2 - 1) = 5(2^2 - 1) = 5 \cdot 3 = 15 \]

Applying the theorem leads to the same result:

\[ \lim_{x \to 2} 5(x^2 - 1) = 5 \cdot \lim_{x \to 2} (x^2 - 1) = 5 \cdot 3 = 15 \]

The final value is unchanged.

Proof

We distinguish two cases: \( k > 0 \) and \( k < 0 \).

1] Case \( k > 0 \)

Assume

\[ \lim_{x \to x_0} f(x) = l \]

By the \( \varepsilon \)-definition of the limit, for every \( \varepsilon > 0 \) there exists a neighborhood of \( x_0 \) such that

\[ l - \frac{\varepsilon}{k} < f(x) < l + \frac{\varepsilon}{k} \]

Multiplying by \( k \), and noting that \( k > 0 \) preserves the inequality signs:

\[ k l - \varepsilon < k f(x) < k l + \varepsilon \]

This is precisely the definition of

\[ \lim_{x \to a} [k f(x)] = k l \]

2] Case \( k < 0 \)

If \( k < 0 \), then \( -k > 0 \). Applying the previous result to \( -k \):

\[ \lim_{x \to a} [(-k) f(x)] = (-k) l \]

By the definition of the limit, for every \( \varepsilon > 0 \) there exists a neighborhood of \( x_0 \) such that

\[ -k l - \varepsilon < (-k) f(x) < -k l + \varepsilon \]

Multiplying by \( -1 \) reverses the inequality signs:

\[ k l + \varepsilon > k f(x) > k l - \varepsilon \]

Rewriting:

\[ k l - \varepsilon < k f(x) < k l + \varepsilon \]

Hence,

\[ \lim_{x \to a} [k f(x)] = k l \]

In conclusion,

\[ \lim_{x \to a} [k f(x)] = k \cdot \lim_{x \to a} f(x) \]

As required.

When the Functions Do Not Both Have Finite Limits

The product rule applies directly when both functions admit finite limits.

The analysis becomes more delicate when zero or infinity appears. In particular, the expression \( 0 \cdot \infty \) is an indeterminate form.

If at least one function does not admit a finite limit, different outcomes are possible.

	ℓ > 0	ℓ < 0	0	+∞	-∞
m > 0	m·ℓ	m·ℓ	0	+∞	-∞
m < 0	m·ℓ	m·ℓ	0	-∞	+∞
0	0	0	0	?	?
+∞	+∞	-∞	?	+∞	-∞
-∞	-∞	+∞	?	-∞	+∞

The cells marked with a question mark correspond to indeterminate forms, which require more advanced analytical tools.

In general, when finite limits are not guaranteed, the product rule cannot be applied mechanically.

Practical example

Consider

\[ \lim_{x \to +\infty} x \cdot e^{-x} \]

We have

\[ \lim_{x \to +\infty} x = \infty \]

\[ \lim_{x \to +\infty} e^{-x} = 0 \]

Thus,

\[ \infty \cdot 0 \]

This is an indeterminate form, so the product rule does not apply directly.

Evaluation strategy

Rewrite the product as a quotient:

\[ \lim_{x \to +\infty} x e^{-x} = \lim_{x \to +\infty} \frac{x}{e^x} \]

We obtain the indeterminate form

\[ \frac{\infty}{\infty} \]

Since the exponential function grows faster than any linear function,

\[ \lim_{x \to +\infty} \frac{x}{e^x} = 0 \]

Hence, the original limit equals zero.