Quotient Law for Limits

Assume that two functions $ f(x) $ and $ g(x) $ have finite limits at a point $ x_0 $: $$ \lim_{x \to x_0} f(x) = l $$ $$ \lim_{x \to x_0} g(x) = m $$ If $ m \ne 0 $, then the limit of their quotient equals the quotient of their limits: $$ \lim_{x \to x_0} \frac{f(x)}{g(x)} = \frac{\lim_{x \to x_0} f(x)}{\lim_{x \to x_0} g(x)} = \frac{l}{m} $$

In short, the limit operation preserves division, provided the denominator does not approach zero.

The requirement $ m \neq 0 $ is not optional. If the denominator approaches zero, the quotient may fail to have a finite limit, or it may lead to an indeterminate form.

Note. The fundamental issue is simple: division by zero is undefined. This is precisely why we must study indeterminate forms such as $ \frac{0}{0} $, $ \frac{\infty}{\infty} $, and similar expressions. These cases cannot be resolved by direct substitution and require more refined analytical tools.

Worked Example
Proof

Worked Example

Compute the limit

$$ \lim_{x \to x_0} \frac{2x+1}{x+3} $$

Let

$$ f(x) = 2x + 1 $$ $$ g(x) = x + 3 $$

Now evaluate the limits of both functions as $ x \to 1 $:

$$ \lim_{x \to 1} f(x) = 2(1) + 1 = 3 $$ $$ \lim_{x \to 1} g(x) = 1 + 3 = 4 $$

Since the limit of the denominator is $ 4 \neq 0 $, the Quotient Law applies:

$$ \lim_{x \to 1} \frac{2x+1}{x+3} = \frac{3}{4} $$

This result agrees with the value of the function at $ x = 1 $ because both the numerator and the denominator are polynomials, and polynomials are continuous everywhere.

Note. Why is the condition $ m \neq 0 $ necessary? Consider the limit $$ \lim_{x \to 2} \frac{x-2}{x-2}. $$ For every real number where it is defined, the expression $ \frac{x-2}{x-2} = 1 $ Therefore, $$ \lim_{x \to 2} \frac{x-2}{x-2} = 1 $$ However, the limit of the denominator as $ x \to 2 $ is $$ \lim_{x \to 2} (x-2) = 0 $$ If we were to apply the Quotient Law mechanically, we would write $$ \lim_{x \to 2} \frac{x-2}{x-2} = \frac{\lim_{x \to 2} (x-2)}{\lim_{x \to 2} (x-2)} = \frac{0}{0} $$ which is an indeterminate form. This example makes it clear that the restriction on the denominator is a genuine mathematical necessity, not a mere technical detail.

Proof

Rewrite the quotient as a product:

$$ \frac{f(x)}{g(x)} = f(x) \cdot \frac{1}{g(x)} $$

$$ \lim_{x \to x_0} g(x) = m \neq 0 $$

then, by the limit law for reciprocal functions,

$$ \lim_{x \to x_0} \frac{1}{g(x)} = \frac{1}{m} $$

Now apply the product limit law:

$$ \lim_{x \to x_0} \left( f(x) \cdot \frac{1}{g(x)} \right) = \left( \lim_{x \to x_0} f(x) \right) \cdot \left( \lim_{x \to x_0} \frac{1}{g(x)} \right) $$

Substituting the known limits yields

$$ l \cdot \frac{1}{m} = \frac{l}{m} $$

This completes the proof.

And so on.