Mathematics · Ch 7 — Applications of Differential Calculus

Indeterminate Forms 0/0, Infinity/Infinity, 0×Infinity, Infinity−Infinity

7.5.3

Indeterminate Forms 0/0, Infinity/Infinity, 0×Infinity, Infinity−Infinity

Direct $\tfrac00$ and $\tfrac{\infty}{\infty}$ applications. For a ratio that is already $\tfrac00$ or $\tfrac{\infty}{\infty}$ at the point in question, differentiate numerator and denominator separately (not as a quotient — do not use the quotient rule) and take the limit of the new ratio; reapply if it is again indeterminate.

Watch out

l'Hôpital's Rule only ever applies to a ratio that is genuinely indeterminate at that step. A famous trap: $\displaystyle\lim_{x\to0^+}\frac{\sin x}{x^2}$ is $\tfrac00$ , and one round of l'Hôpital gives $\displaystyle\lim_{x\to0^+}\frac{\cos x}{2x}=\frac{1}{0^+}=+\infty$ — this is the correct final answer precisely because $\tfrac{\cos x}{2x}$ is no longer an indeterminate form, so the limit is read off directly; applying l'Hôpital again to it (mistakenly treating $\tfrac10$ as if it were still indeterminate) would give a wrong result.

Reducing $0\times\infty$ to $\tfrac00$ or $\tfrac{\infty}{\infty}$ . If $\displaystyle\lim f(x)=0$ and $\displaystyle\lim g(x)=\infty$ , rewrite the product as $f(x)\,g(x)=\dfrac{f(x)}{1/g(x)}$ (a $\tfrac00$ form) or $\dfrac{g(x)}{1/f(x)}$ (a $\tfrac{\infty}{\infty}$ form), whichever differentiates more simply, then apply l'Hôpital. …

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