First check that the quotient tends to the indeterminate form 00 or. Under the usual differentiability conditions, with non-zero denominator derivative near the limit, evaluate the limit of the quotient of derivatives when that limit exists. lim_x to a f(x)g(x)=lim_x to a f (x)g (x). Differentiate numerator and denominator separately, not with the quotient rule. Recheck the indeterminate form before repeating. Examples lim_x 0 ex-1-xx2=lim_x 0 ex-12x=lim_x 0 ex2= 12. lim_x to infinity ln xx=lim_x to infinity 1x1=0. By Maclaurin, sin x=x- x36+. s, so lim_x 0 sin x-xx3=- 16. Retain the first non-cancelling term. Direct substitution or algebra is preferable when it already gives a defined limit. A form such as 10 does not permit l’Hôpital’s rule.

First check that the quotient tends to the indeterminate form \(\dfrac00\) or \(\dfrac\infty\infty\). Under the usual differentiability conditions, with non-zero denominator derivative near the limit, evaluate the limit of the quotient of derivatives when that limit exists.

\(\lim_{x\to a}\dfrac{\mathrm{f}(x)}{\mathrm{g}(x)}=\lim_{x\to a}\dfrac{\mathrm{f}^{\prime}(x)}{\mathrm{g}^{\prime}(x)}\). Differentiate numerator and denominator separately, not with the quotient rule. Recheck the indeterminate form before repeating.

\(\lim_{x\to0}\dfrac{e^x-1-x}{x^2}=\lim_{x\to0}\dfrac{e^x-1}{2x}=\lim_{x\to0}\dfrac{e^x}{2}=\dfrac12\).

\(\lim_{x\to\infty}\dfrac{\ln x}{x}=\lim_{x\to\infty}\dfrac{\dfrac1x}{1}=0\).

By Maclaurin, \(\sin x=x-\dfrac{x^3}{6}+\cdots\), so \(\lim_{x\to0}\dfrac{\sin x-x}{x^3}=-\dfrac16\). Retain the first non-cancelling term.

Direct substitution or algebra is preferable when it already gives a defined limit. A form such as \(\dfrac10\) does not permit l’Hôpital’s rule.

Limits and Continuity

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Limits by l’Hôpital’s Rule and Series

Limits by l’Hôpital’s Rule and Series