L'Hôpital's Rule

109. L’Hôpital’s Rule

Evaluating limits that result in a indeterminite form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$

If ${lim}_{𝑥 \to 𝑎} 𝑓 (𝑥) = 0$ and ${lim}_{𝑥 \to 𝑎} 𝑔 (𝑥) = 0$ (or both go to $\pm \infty$ ), and $𝑓 (𝑥)$ and $𝑔 (𝑥)$ are differentiable near $𝑎$ , then:

lim_{𝑥 \to 𝑎} \frac{𝑓 (𝑥)}{𝑔 (𝑥)} = lim_{𝑥 \to 𝑎} \frac{𝑓^{'} (𝑥)}{𝑔^{'} (𝑥)}

Example

Consider:

lim_{𝑥 \to 0} \frac{1 - cos (𝑥)}{𝑥^{2}}

Step 1: Direct Substitution

Substituting $𝑥 = 0$ :

\frac{1 - cos (0)}{0^{2}} = \frac{1 - 1}{0} = \frac{0}{0}

Since this is an indeterminate form, we apply L’Hôpital’s Rule.

Step 2: First Application of L’Hôpital’s Rule

Differentiate the numerator and denominator:

Numerator: $𝑓 (𝑥) = 1 - cos (𝑥) \Rightarrow 𝑓^{'} (𝑥) = sin (𝑥)$
Denominator: $𝑔 (𝑥) = 𝑥^{2} \Rightarrow 𝑔^{'} (𝑥) = 2 𝑥$

Thus, applying L’Hôpital’s Rule:

lim_{𝑥 \to 0} \frac{1 - cos (𝑥)}{𝑥^{2}} = lim_{𝑥 \to 0} \frac{sin (𝑥)}{2 𝑥}

Step 3: Second Check for Indeterminate Form

Substituting $𝑥 = 0$ :

\frac{sin (0)}{2 (0)} = \frac{0}{0}

Since this is still an indeterminate form, we apply L’Hôpital’s Rule again.

Step 4: Second Application of L’Hôpital’s Rule

Differentiate again:

Numerator: $𝑓^{'} (𝑥) = sin (𝑥) \Rightarrow 𝑓^{″} (𝑥) = cos (𝑥)$
Denominator: $𝑔^{'} (𝑥) = 2 𝑥 \Rightarrow 𝑔^{″} (𝑥) = 2$

Applying L’Hôpital’s Rule again:

lim_{𝑥 \to 0} \frac{sin (𝑥)}{2 𝑥} = lim_{𝑥 \to 0} \frac{cos (𝑥)}{2}

Step 5: Evaluate the Limit

Now, substituting $𝑥 = 0$ :

\frac{cos (0)}{2} = \frac{1}{2}

Final Answer:

lim_{𝑥 \to 0} \frac{1 - cos (𝑥)}{𝑥^{2}} = \frac{1}{2}