What if the indeterminate form is $0 \cdot \infty$?

Rewrite the product $f(x)g(x)$ as a quotient $\frac{f(x)}{1/g(x)}$ or $\frac{g(x)}{1/f(x)}$ to force a $0/0$ or $\infty/\infty$ form, then apply the rule.

What if the form is $\infty - \infty$?

Combine the terms algebraically into a single fraction using a common denominator to achieve a $0/0$ or $\infty/\infty$ form.

What if the form is $1^\infty$, $0^0$, or $\infty^0$?

Set the limit equal to $y$. Take the natural logarithm of both sides to convert the exponent into a product, rewrite as a quotient, apply L'Hopital's rule to find $\lim \ln(y)$, and exponentiate the result to find $\lim y$.

How to apply L'Hopital's rule

L'Hopital's rule evaluates limits of indeterminate forms by differentiating the numerator and denominator independently. It applies strictly to limits resulting in $0/0$ or $\pm\infty/\infty$ .

The setup

Identify the limit $\lim_{x o c} \frac{f(x)}{g(x)}$ . Substitute $c$ into $f(x)$ and $g(x)$ to verify the expression evaluates exactly to the indeterminate form $0/0$ or $\pm\infty/\infty$ .

The steps

Differentiate the numerator $f(x)$ to find $f'(x)$ . 2. Differentiate the denominator $g(x)$ to find $g'(x)$ . Do not use the quotient rule. 3. Form the new limit $\lim_{x o c} \frac{f'(x)}{g'(x)}$ . 4. Substitute $c$ into the new limit. 5. If the new limit yields another indeterminate form, repeat steps 1-4.

Checking the result

Substitute a test value extremely close to $c$ (e.g., $c + 0.001$ ) into the original function $\frac{f(x)}{g(x)}$ . The numerical output must closely approximate your calculated limit.

Common errors

Applying the quotient rule to $\frac{f(x)}{g(x)}$ instead of taking derivatives independently. Applying the rule when the limit is not indeterminate (e.g., $1/0$ or $0/1$ ), which yields incorrect results.

Worked example

Evaluate $\lim_{x o 0} \frac{\sin(x) - x}{x^3}$ .

Substitute $x = 0$ : $\frac{\sin(0) - 0}{0^3} = \frac{0}{0}$ . The form is indeterminate. Apply L'Hopital's rule by differentiating numerator and denominator: $\lim_{x o 0} \frac{\cos(x) - 1}{3x^2}$ . Substitute $x = 0$ : $\frac{\cos(0) - 1}{0} = \frac{0}{0}$ . The form is still indeterminate. Apply L'Hopital's rule a second time: $\lim_{x o 0} \frac{-\sin(x)}{6x}$ . Substitute $x = 0$ : $\frac{-\sin(0)}{0} = \frac{0}{0}$ . Apply L'Hopital's rule a third time: $\lim_{x o 0} \frac{-\cos(x)}{6}$ . Substitute $x = 0$ : $\frac{-\cos(0)}{6} = -\frac{1}{6}$ . The limit is $-1/6$ .

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References: Stewart Calculus, 8th Edition · OpenStax Calculus Volume 1 · Khan Academy: Applications of derivatives