A question about the L'Hopital's rule

[Artusartos]

This is how L'Hopital's Rule is defined in our textbook:


"Let s signify $a, a^+, a^-, \infty or - \infty$ and suppose f and g are differentiable functions for which hte following limit exists:

$lim_{x \rightarrow s} \frac{f'(x)}{g'(x)} = L$..................(1)

If

$lim_{x \rightarrow s} f(x) = lim_{x \rightarrow s} g(x) = 0$................(2)

or if

$lim_{x \rightarrow s} |g(x)| = + \infty$.....................(3)

then

$lim_{x \rightarrow s} \frac{f(x)}{g(x)} = L$....................(4)


Here is the part that I was having trouble with...

Note that the hypthesis (1) includes some implicit assumptions: f and g must be defined and differentiable "near" s and g'(x) must be nonzero "near" s. For example, if $lim_{x \rightarrow a^+} \frac{f'(x)}{g'(x)}$ exists, then there g' is nonzero. The requirement that g' be nonzero is crucial; see Exercise 30.7.


This is what exercise 30.7 is:

This example is taken from [38] and is due to Otto Stolz, Math, annalen 15 (1879), 556-559. The requirement in Theorem 30.2 [the one that I wrote above] that $g'(x) \not= 0$ for x "near" s is important. In a careless application of L'Hopital's rule in which the zeros of g' "cancel" the zeros of f' erroneous results can be obtained. For x in R, let

f(x) = x + cosxsinx and $g(x)= e^{sinx}(x+cosxsinx)$

$\frac{f'(x)}{g'(x)} = \frac{2e^{-sinx}cosx}{2cosx+f(x)}$ if cosx is not zero adn x>3.

Show that $lim_{x \rightarrow \infty} \frac{2e^{-sinx}cosx}{2cosx+f(x)} = 0$ and yet the limit $lim_{x \rightarrow \infty} \frac{f(x)}{g(x)}$ does not exist.

I'm a bit confused about this, because I don't think I understood that theorem correctly...so can anybody clarify it for me?

For example, with this exercise...when I graphed f(x) and g(x), they both looked continuous and smooth...so obviously they were differentiable. But I could not understand how g'(x) is zero "near" infinity, because (as I said), I'm having trouble understanding the theorem...

 

[MarneMath]

Since no one else answered. I'm a going to ask for some clarification. Are you having trouble understanding why this rule is true? Or are you having trouble seeing what this problem is trying to get you realize?