M171 Calculus I · Unit 3: Implicit, limits & extrema · on Exam 3
§2.8 L'Hôpital's rule and limits at infinity
Study notes for L'Hôpital's rule and limits at infinity in M171 Calculus I at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in Mathub.
Big ideas
- L'Hôpital: if $\lim \frac{f(x)}{g(x)}$ has the form $\frac00$ or $\frac{\infty}{\infty}$, then it equals $\lim\frac{f'(x)}{g'(x)}$ (when that limit exists). Differentiate top and bottom separately, not with the quotient rule. Always verify the form first.
- Limits at infinity describe end behaviour and horizontal asymptotes. For rational functions compare degrees: numerator lower $\to 0$; equal $\to$ ratio of leading coefficients; numerator higher $\to \pm\infty$.
- Dominating terms: as $x\to\infty$, exponentials $e^{x}$ beat powers $x^n$, which beat logarithms $\ln x$. E.g. $\frac{x^{100}}{e^x}\to0$ and $\frac{\ln x}{\sqrt x}\to0$.
Key formulas
L'Hôpital's rule
$$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}\quad\text{for forms } \tfrac00,\ \tfrac\infty\infty$$
Rational end behaviour
$$\lim_{x\to\infty}\frac{3x^2 - x}{5x^2 + 4} = \frac35$$
Hierarchy
$$\ln x \ll x^p \ll e^{x}\quad (x\to\infty)$$
Worked example
Evaluate $\lim_{x\to0}\dfrac{\sin 5x}{x}$ and $\lim_{x\to\infty}\dfrac{x^2}{e^x}$.
Show the solution
First: form $\frac00$, so $\lim \frac{5\cos 5x}{1} = 5$. Second: form $\frac\infty\infty$; apply twice: $\lim\frac{2x}{e^x} = \lim\frac{2}{e^x} = 0$ (exponential dominates).
Common mistakes
- Applying L'Hôpital to a limit that is not $\frac00$ or $\frac\infty\infty$ (e.g. $\lim_{x\to0}\frac{\cos x}{x}$ is a vertical asymptote, not L'Hôpital).
- Using the quotient rule instead of differentiating numerator and denominator separately.
Exam tipWrite the form in words above the equals sign, "(0/0)", each time you apply the rule.
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