Lhopitals Rule

Understanding L'Hopital's Rule

When direct substitution in $\lim_{x\to a} \frac{f(x)}{g(x)}$ yields $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, we face an indeterminate form—the limit may or may not exist, and simple substitution cannot tell you which. L'Hopital's Rule provides a systematic solution: differentiate the numerator and denominator independently and evaluate the new ratio's limit.

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

provided the right-hand limit exists. This works because near the point in question, both functions behave like their tangent line approximations; comparing slopes reveals true limiting behavior.

Key Steps

1. Verify the form: Confirm substitution gives exactly $\frac{0}{0}$ or $\frac{\infty}{\infty}$
2. Differentiate separately: Apply derivative to top and bottom independently—this is not the quotient rule
3. Re-check after each step: If the new ratio remains indeterminate, apply the rule again; if not, substitute directly

Common Examples

For $\lim_{x\to 0} \dfrac{\sin x}{x}$: substitution yields $\frac{0}{0}$. Differentiating gives $\lim_{x\to 0}\dfrac{\cos x}{1} = 1$.

For $\lim_{x\to\infty} \dfrac{x^2}{e^x}$: the form $\frac{\infty}{\infty}$ requires applying the rule twice. After the first application: $\lim_{x\to\infty}\dfrac{2x}{e^x}$ remains $\frac{\infty}{\infty}$. After the second: $\lim_{x\to\infty}\dfrac{2}{e^x} = 0$.

Extending Beyond Standard Forms

Other indeterminate forms—$0\cdot\infty$, $\infty-\infty$, $1^\infty$, $0^0$, $\infty^0$—can often be algebraically rewritten into $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then solved using the rule.

Common Pitfall

Failing to verify that the form is genuinely indeterminate before applying the rule, or forgetting to re-verify after each differentiation step, leads to incorrect answers. Always check the form first.