🤔 What is L'Hôpital's Rule?
L'Hôpital's Rule states that if the limit of $f(x)/g(x)$ as $x$ approaches $c$ results in an indeterminate form (0/0 or ∞/∞), then:
$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$
provided the limit on the right exists.
📝 Steps to Apply L'Hôpital's Rule
- Verify Indeterminate Form: Check if the limit results in 0/0 or ∞/∞.
- Differentiate: Find the derivatives of the numerator $f'(x)$ and the denominator $g'(x)$.
- Evaluate the New Limit: Compute the limit of $f'(x)/g'(x)$ as $x$ approaches $c$.
- Repeat if Necessary: If the new limit is still an indeterminate form, repeat steps 2 and 3.
🚀 Example 1: 0/0 Form
Evaluate: $\lim_{x \to 0} \frac{\sin(x)}{x}$
- Verify Indeterminate Form: As $x \to 0$, $\sin(x) \to 0$ and $x \to 0$, so we have 0/0.
- Differentiate:
- $f(x) = \sin(x) \implies f'(x) = \cos(x)$
- $g(x) = x \implies g'(x) = 1$
- Evaluate the New Limit:
$\lim_{x \to 0} \frac{\cos(x)}{1} = \frac{\cos(0)}{1} = \frac{1}{1} = 1$
Therefore, $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$
♾️ Example 2: ∞/∞ Form
Evaluate: $\lim_{x \to \infty} \frac{x^2}{e^x}$
- Verify Indeterminate Form: As $x \to \infty$, $x^2 \to \infty$ and $e^x \to \infty$, so we have ∞/∞.
- Differentiate (First Application):
- $f(x) = x^2 \implies f'(x) = 2x$
- $g(x) = e^x \implies g'(x) = e^x$
$\lim_{x \to \infty} \frac{2x}{e^x}$
- Verify Indeterminate Form Again: As $x \to \infty$, $2x \to \infty$ and $e^x \to \infty$, so we still have ∞/∞.
- Differentiate (Second Application):
- $f'(x) = 2x \implies f''(x) = 2$
- $g'(x) = e^x \implies g''(x) = e^x$
$\lim_{x \to \infty} \frac{2}{e^x}$
- Evaluate the New Limit:
$\lim_{x \to \infty} \frac{2}{e^x} = 0$
Therefore, $\lim_{x \to \infty} \frac{x^2}{e^x} = 0$
⚠️ Important Notes
- L'Hôpital's Rule only applies to indeterminate forms 0/0 and ∞/∞.
- Ensure that both $f(x)$ and $g(x)$ are differentiable.
- Sometimes, you may need to apply the rule multiple times.