4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

A quotient such as $\dfrac{0}{0}$ does not equal zero, and $\dfrac{\infty}{\infty}$ is not a numerical fraction. These symbols label an indeterminate form: direct substitution has not revealed the limit, because different functions with the same apparent form can approach different values.

Enduring Understanding — LIM-4: L’Hospital’s Rule allows us to determine the limits of some indeterminate forms.

The learning objective is LIM-4.A: Determine limits of functions that result in indeterminate forms. Essential knowledge LIM-4.A.1 identifies $\dfrac{0}{0}$ and $\dfrac{\infty}{\infty}$ as indeterminate forms. Essential knowledge LIM-4.A.2 connects these forms with L’Hospital’s Rule as a possible method.

First diagnose the form

When ordinary limit laws apply and the denominator approaches a nonzero number, a quotient can be evaluated by dividing the separate limits. But if the denominator approaches zero, that quotient rule cannot be used:

$$ \lim_{x\to a}\frac{f(x)}{g(x)} \neq \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)} \qquad\text{when}\qquad \lim_{x\to a}g(x)=0. $$

The correct first move is direct substitution, not immediate differentiation.

1. Compute the limiting behavior of the numerator.
2. Compute the limiting behavior of the denominator.
3. Apply L’Hospital’s Rule only if the result is $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$.
4. Differentiate the numerator and denominator separately.
5. Reevaluate the new limit.

For example,

$$ \lim_{x\to 1}\frac{x^2-1}{x-1} $$

produces

$$ \frac{1^2-1}{1-1}=\frac{0}{0}. $$

This does not mean the limit equals $\dfrac{0}{0}$. It means direct substitution is inconclusive. In this particular case, factoring is enough:

$$ \frac{x^2-1}{x-1}

\frac{(x-1)(x+1)}{x-1}

x+1 \qquad (x\ne1), $$

so

$$ \lim_{x\to1}\frac{x^2-1}{x-1}

\lim_{x\to1}(x+1)=2. $$

L’Hospital’s Rule

L’Hospital’s Rule says that, under the appropriate differentiability and existence conditions, if

$$ \lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0 $$

or both functions approach infinity, then

$$ \lim_{x\to a}\frac{f(x)}{g(x)}

\lim_{x\to a}\frac{f'(x)}{g'(x)}, $$

provided the limit on the right exists or approaches infinity in a suitable way.

The rule compares rates of approach. It does not say that a quotient of limits may be replaced mechanically by a quotient of derivatives in every problem. The indeterminate form must be verified first.

The $\dfrac{0}{0}$ form

Consider

$$ \lim_{x\to0}\frac{\sin x}{x}. $$

Direct substitution gives

$$ \frac{\sin 0}{0}=\frac{0}{0}, $$

so L’Hospital’s Rule is eligible:

$$ \lim_{x\to0}\frac{\sin x}{x}

\lim_{x\to0}\frac{\cos x}{1}

\frac{\cos0}{1}

$$

The same limit can be established by other calculus methods; L’Hospital’s Rule is useful because differentiation removes the indeterminate quotient.

The $\dfrac{\infty}{\infty}$ form

Now examine

$$ \lim_{x\to\infty}\frac{3x^2+1}{2x^2-5}. $$

As $x\to\infty$, both the numerator and denominator approach infinity, so the expression has the form $\dfrac{\infty}{\infty}$. Applying the rule gives

$$ \lim_{x\to\infty}\frac{6x}{4x}

\lim_{x\to\infty}\frac{3}{2}

\frac{3}{2}. $$

Here the rule exposes the dominant growth rates of the two functions.

A scope and notation warning

Misconception — “The limit equals $\dfrac{0}{0}$.” Writing

$$ \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{0}{0} $$

is incorrect. Instead write: “Direct substitution gives the indeterminate form $\dfrac{0}{0}$,” and then continue with an appropriate method.

For AP Calculus AB and BC, $\dfrac{0}{0}$ and $\dfrac{\infty}{\infty}$ are recognized as indeterminate forms, but L’Hospital’s Rule is not required as an AP-exam method. Treat it as an enrichment technique unless a particular course or teacher explicitly assesses it. Other forms, including $\infty-\infty$, are outside the assessed AP Calculus AB and BC scope described here.

When L’Hospital’s Rule is used in a setting that permits it, the reasoning must include the diagnosis:

“Because the numerator and denominator both approach $0$, the expression has the indeterminate form $\dfrac{0}{0}$; therefore L’Hospital’s Rule may be applied.”

This explicitly demonstrates Skill 3.D: Use mathematical reasoning to solve problems by checking the condition before executing the procedure.

Retrieval check

For

$$ \lim_{x\to2}\frac{x^2-4}{x^2-3x+2}, $$

direct substitution gives $\dfrac{0}{0}$. Is that the answer? No: it is only a diagnosis. Factoring produces

$$ \frac{(x-2)(x+2)}{(x-2)(x-1)}

\frac{x+2}{x-1}, $$

so the limit is

$$ \boxed{4}. $$

The essential habit is: verify the indeterminate form first, then choose a valid method.