Editing L'Hôpital's rule (section)

=== 4. Limit of derivatives does not exist ===
The requirement that the limit <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> exists is essential; if it does not exist, the original limit <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> may nevertheless exist. Indeed, as <math>x</math> approaches <math>c</math>, the functions <math>f</math> or <math>g</math> may exhibit many oscillations of small amplitude but steep slope, which do not affect <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> but do prevent the convergence of <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math>. 

For example, if <math>f(x)=x+\sin(x)</math>, <math>g(x)=x</math> and <math>c=\infty</math>, then <math display="block">\frac{f'(x)}{g'(x)}=\frac{1+\cos(x)}{1},</math>which does not approach a limit since cosine oscillates infinitely between {{math|1}} and {{math|−1}}. But the ratio of the original functions does approach a limit, since the amplitude of the oscillations of <math>f</math> becomes small relative to <math>g</math>:

:<math>\lim_{x\to\infty}\frac{f(x)}{g(x)} 
= \lim_{x\to\infty}\left(\frac{x+\sin(x)}{x}\right) 
= \lim_{x\to\infty}\left(1+\frac{\sin(x)}{x}\right) 
= 1+0 = 1. </math>

In a case such as this, all that can be concluded is that

: <math> \liminf_{x \to c} \frac{f'(x)}{g'(x)} \leq \liminf_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f'(x)}{g'(x)} ,</math>

so that if the limit of <math display="inline">\frac{f}{g}
</math> exists, then it must lie between the inferior and superior limits of <math display="inline">\frac{f'}{g'}
</math> .  In the example, 1 does indeed lie between 0 and 2.)

Note also that by the [[Contraposition|contrapositive]] form of the Rule, if <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> does not exist, then <math>\lim_{x\to c}\frac{f'(x)}{g'(x)}</math> also does not exist.