I'm wondering if for some function whereby the numerator and denominator goes to $0$ and L'Hopital Rule can be applied to find the limit of the function, does it necessarily imply that, $\lim (\frac{f(x)}{g(x)} - \frac{f'(x)}{g'(x)}) = 0$.

I'm asking this question because I came across this question that asked me to find the limit of the following expression:

$$\lim_{x \rightarrow 0}(\frac{e^x+e^{-x}-2}{1-\cos2x}-\frac{e^x-e^{-x}}{2\sin2x})$$

This limit goes to zero. It may be possible to combine the fractions and use L'Hopital's on the entire fraction, but that is tedious to do. Upon closer inspection, the second term is just what you'd get if you did L'Hopital on the first expression. Since the limits of both terms should approach the same value, shouldn't they 'cancel out', in a sense?

I'm wondering if $\lim (\frac{f(x)}{g(x)} - \frac{f'(x)}{g'(x)}) = 0$ is always true for functions which are of indeterminate form. I've been playing around and I've yet to find a counterexample.

  • 1
    $\begingroup$ You could rewrite your limit as $-\lim\frac{g}{g^\prime}\frac{d}{dx}\frac{f}{g}$ to see what conditions ensure it's $0$. $\endgroup$ – J.G. Nov 25 '19 at 8:15

Any time $\frac{f'(x)}{g'(x)}$ has finite limit, it will necessarily be true because both $\frac{f(x)}{g(x)}$ and $\frac{f'(x)}{g'(x)}$ go to the same limit. And the limit of a difference is equal to the difference of the limits, as long as the limits are finite.

However, for indeterminate forms that happen to go to infinity there is no reason that this should be true. For instance, take $$ \lim_{x\to 0}\left(\frac{x}{x^3} - \frac{1}{3x^2}\right) = \lim_{x\to 0}\frac{2}{3x^2}\neq 0 $$ And similarily, if $\frac{f'(x)}{g'(x)}$ fails to have a limit at all, there is no reason the difference should go to $0$.

  • $\begingroup$ The fraction $\frac{f(x)}{g(x)}=\frac12$ has finite limit but $\frac{f'(x)}{g'(x)}=\frac00$. $\endgroup$ – user Nov 25 '19 at 8:25
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    $\begingroup$ I've noticed you have added "assuming both exist" but maybe it should be more clear if stated in the spirit of l'Hopital's rule. That is when $\frac{f(x)}{g(x)}$ is an indeterminate form and $\frac{f'(x)}{g'(x)} \to L$ then $\lim_{x\to x_0} \left(\frac{f(x)}{g(x)} - \frac{f'(x)}{g'(x)}\right)=0$. $\endgroup$ – user Nov 25 '19 at 8:29
  • $\begingroup$ @user It's been a while, but I was just reminded of this answer due to an upvote. I have changed the wording to be more in the actual spirit of l'Hopital. As in in line with the statement of the theorem, not what it's used for by calculus students. $\endgroup$ – Arthur yesterday

By l'Hopital'rule when $\frac{f(x)}{g(x)}$ is an indeterminate form and

$$\lim_{x\to x_0} \frac{f'(x)}{g'(x)} =L$$

we necessarly have that

$$\lim_{x\to x_0} \left(\frac{f(x)}{g(x)} - \frac{f'(x)}{g'(x)}\right)=0$$

otherwise when $\lim_{x\to x_0} \frac{f'(x)}{g'(x)}$ doesn't exist it is not true in general.


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