Understanding L’Hôpital’s Rule with Examples

L’Hôpital’s Rule is a powerful method in calculus that helps us evaluate limits that result in indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

In this blog post, we will explore what L’Hôpital’s Rule is, when to use it, and how to apply it effectively with examples.

What is L’Hôpital’s Rule?

L’Hôpital’s Rule states that if: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \quad \text{or} \quad \frac{\infty}{\infty},\] then:\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},\] provided that the limit on the right-hand side exists. This means we can differentiate the numerator and denominator separately and then take the limit again.

When to Use L’Hôpital’s Rule

L’Hôpital’s Rule applies only when:

1. The limit is of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
2. The functions \( f(x) \) and \( g(x) \) are differentiable near \( x = a \).

If the limit is not indeterminate, L’Hôpital’s Rule cannot be applied.

How to Apply L’Hôpital’s Rule

Here’s a step-by-step guide to applying L’Hôpital’s Rule:

1. Check the Indeterminate Form: Verify that the limit is \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
2. Differentiate the Numerator and Denominator: Compute \( f'(x) \) and \( g'(x) \).
3. Take the Limit Again: Evaluate \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \).
4. Repeat if Necessary: If the new limit is still indeterminate, apply L’Hôpital’s Rule again.

When Not to Use L’Hôpital’s Rule

• If the limit is not indeterminate (e.g., \( \frac{1}{0} \) or \( \frac{\infty}{0} )\), L’Hôpital’s Rule does not apply.
• If differentiating the numerator and denominator makes the problem more complicated, consider alternative methods (e.g., factoring, rationalizing, or using known limits).

Example of L’Hôpital’s Rule

Example 1:

Evaluate the following limit:
\[
\lim_{x \to 0} \frac{\sin(x)}{x}
\]

Solution:

1. Check the form: At \( x = 0 \), \( \sin(0) = 0 \) and \( x = 0 \), so the limit is \( \frac{0}{0} \).
2. Apply L’Hôpital’s Rule:
   \[
   \lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\cos(x)}{1}
   \]
3. Evaluate the new limit:
   \[
   \lim_{x \to 0} \frac{\cos(x)}{1} = \cos(0) = 1
   \]
   So, solution is \( \fbox{$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$} \)

Example 2:

Evaluate the following limit:
\[
\lim_{x \to \infty} \frac{x^2}{e^x}
\]

Solution:

1. Check the form: As \( x \to \infty \), \( x^2 \to \infty \) and \( e^x \to \infty \), so the limit is \( \frac{\infty}{\infty} \).
2. Apply L’Hôpital’s Rule:
   \[
   \lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x}
   \]
3. The new limit is still \( \frac{\infty}{\infty} \), so apply L’Hôpital’s Rule again:
   \[
   \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x}
   \]
4. Evaluate the new limit:
   \[
   \lim_{x \to \infty} \frac{2}{e^x} = 0
   \]
   So, solution is \( \fbox{$\lim_{x \to \infty} \frac{x^2}{e^x} = 0$} \)