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Mathematics · Ch 18 — Differentiation

Brief Idea of L'Hospital's Rule

18.2.8

Brief Idea of L'Hospital's Rule

Brief idea of L'Hospital's Rule. Consider two functions f(x)f(x) and g(x)g(x). If lim⁡x→af(x)=0\displaystyle\lim_{x\to a}f(x)=0 and lim⁡x→ag(x)=0\displaystyle\lim_{x\to a}g(x)=0 (a 00\frac00 indeterminate form), and if lim⁡x→af′(x)=p\displaystyle\lim_{x\to a}f'(x)=p and lim⁡x→ag′(x)=q\displaystyle\lim_{x\to a}g'(x)=q where q≠0q\ne0, then

lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)=pq\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}=\dfrac pq

If instead lim⁡x→ag′(x)=0\displaystyle\lim_{x\to a}g'(x)=0 too, then — provided lim⁡x→af′(x)=0\displaystyle\lim_{x\to a}f'(x)=0 as well — the same 00\frac00 situation reappears one derivative down, and lim⁡x→af′(x)g′(x)\displaystyle\lim_{x\to a}\dfrac{f'(x)}{g'(x)} can itself be studied by the same rule (differentiating top and bottom again).

Example 1: lim⁡x→0sin⁡xx2\displaystyle\lim_{x\to0}\dfrac{\sin x}{x^2}. Here f(x)=sin⁡xf(x)=\sin x with lim⁡x→0f(x)=0\lim_{x\to0}f(x)=0, and g(x)=x2g(x)=x^2 with lim⁡x→0g(x)=0\lim_{x\to0}g(x)=0. Then f′(x)=cos⁡xf'(x)=\cos x, so lim⁡x→0f′(x)=cos⁡0=1≠0\lim_{x\to0}f'(x)=\cos0=1\ne0; but g′(x)=2xg'(x)=2x, so lim⁡x→0g′(x)=2(0)=0\lim_{x\to0}g'(x)=2(0)=0. Since lim⁡x→0g′(x)=0\lim_{x\to0}g'(x)=0 (and f′f''s limit is not also 00), L'Hospital's Rule cannot be applied here. …