L'Hospital's rule and indeterminate forms

[PFuser1232]

I want to make sure I understand the conditions required for L'Hospital's Rule to work.
$$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$
If $\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$ exists.
Should $f$ and $g$ be differentiable at $a$? Or just around $a$?
Also, would it work if $g'(a) = 0$?

 

[SteamKing]

I want to make sure I understand the conditions required for L'Hospital's Rule to work.
$$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$
If $\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$ exists.
Should $f$ and $g$ be differentiable at $a$? Or just around $a$?
Also, would it work if $g'(a) = 0$?

It's not clear what you mean by a function being differentiable "just around a".

For univariate functions, a derivative either exists at a given abscissa or it doesn't. I'm not aware of anything in between.

If g'(a) = 0, then it is possible that the first application of L'Hopital's rule leads to another indeterminate form. You can apply L'Hopital's rule serially until either a limit is reached or it becomes clear that no convergent limit will ever be obtained. Then you have to move on.

 

[HallsofIvy]

By "just around a" I believe that MohamedRady97 means "in some neighborhood of a but not necessarily at x= a".
For example, f(x)= x if x< 1, f(x)= 2x- 1 if x> 1 is differentiable "around x= 1" but not at 1.

The answer to his question is "yes, that is correct". The definition of "limit" is such that what happens in a neighborhood of a, NOT at x= a itself, is all that is taken into consideration when taking a limit.

 

[theodoros.mihos]

See this. Theorem have special conditions.

 

[mathwonk]

as suggested above you do not have enough hypotheses to make the theorem true. you also need something like the limits of f and g are both = 0, or both = infinity. look in a book.