Concept preview

What you will be able to do

1. 01Recognise when a limit produces an indeterminate form of type $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ by substituting first, and state that L'Hôpital's Rule applies ONLY to those two forms.
2. 02Apply L'Hôpital's Rule correctly by differentiating the numerator and denominator SEPARATELY (never with the quotient rule), and repeat the process when the new limit is still indeterminate.
3. 03Rewrite disguised indeterminate forms such as $0\cdot\infty$ and $\infty-\infty$ as a single quotient so that L'Hôpital's Rule can be used, and explain intuitively why the rule works.

L'Hôpital's Rule addresses a specific breakdown. Suppose you want $\displaystyle\lim_{x\to c}\dfrac{f(x)}{g(x)}$ and, on substituting $x=c$, you find that both $f(c)=0$ and $g(c)=0$, or that both $f$ and $g$ run off to $\pm\infty$. In either case the ratio is an indeterminate form and substitution alone tells you nothing. The rule states that, under these conditions, the original limit equals the limit of the ratio of derivatives: $\lim_{x\to c}\dfrac{f(x)}{g(x)}=\lim_{x\to c}\dfrac{f'(x)}{g'(x)},$ provided that second limit exists (or is $\pm\infty$). The point $c$ may be a finite number or $\pm\infty$, and the one-sided versions hold too.