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⬅ Back to Differential Calculus Topics

L'Hospital's Rule

$$\text{If } \lim_{x\to a} f(x)=0 \text{ and } \lim_{x\to a} g(x)=0$$
$$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$$

L'Hôpital's Rule — Trig Ratio (CE Board)

Evaluate: $\displaystyle\lim_{x\to 0}\frac{\sin 2x}{\sin 3x}$

Direct substitution gives $\dfrac{0}{0}$ — apply L'Hôpital's Rule:

$\lim_{x\to 0}\dfrac{\sin 2x}{\sin 3x} = \lim_{x\to 0}\dfrac{2\cos 2x}{3\cos 3x}$

Substitute $x = 0$:

$= \dfrac{2\cos 0}{3\cos 0} = \dfrac{2(1)}{3(1)} = \boxed{\dfrac{2}{3}}$

L'Hôpital's Rule — Exponential Limit

Evaluate: $\displaystyle\lim_{x\to 0}\frac{e^x - 1}{x}$

Direct substitution gives $\dfrac{0}{0}$. Apply L'Hôpital's Rule:

$\lim_{x\to 0}\dfrac{e^x - 1}{x} = \lim_{x\to 0}\dfrac{e^x}{1} = e^0 = \boxed{1}$

L'Hôpital Applied Twice — CE Board

Evaluate: $\displaystyle\lim_{x\to 0}\frac{1 - \cos x}{x^2}$

Direct substitution gives $\dfrac{0}{0}$. Apply L'Hôpital's Rule once:

$\lim_{x\to 0}\dfrac{\sin x}{2x}$

Still $\dfrac{0}{0}$ — apply L'Hôpital's Rule again:

$\lim_{x\to 0}\dfrac{\cos x}{2} = \dfrac{\cos 0}{2} = \boxed{\dfrac{1}{2}}$

L'Hôpital's Rule — Infinity Form

Evaluate: $\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}$

As $x\to\infty$, this is the $\dfrac{\infty}{\infty}$ indeterminate form. Apply L'Hôpital's Rule:

$\lim_{x\to\infty}\dfrac{\ln x}{x} = \lim_{x\to\infty}\dfrac{1/x}{1} = \lim_{x\to\infty}\dfrac{1}{x} = \boxed{0}$

This confirms that logarithms grow slower than any power of $x$.

L'Hôpital's Rule — Polynomial (CE Board)

Evaluate: $\displaystyle\lim_{x\to 1}\frac{x^3 - 1}{x - 1}$

Direct substitution gives $\dfrac{0}{0}$. Apply L'Hôpital's Rule:

$\lim_{x\to 1}\dfrac{3x^2}{1} = 3(1)^2 = \boxed{3}$

Verification by factoring: $x^3-1 = (x-1)(x^2+x+1)$, so the limit is $x^2+x+1\big|_{x=1} = 3$ ✓

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