Thread: Need some constructive criticism...(advanced calc proof)

I am unsure of this proof, and I would appreciate any input about whether my reasoning is on the right track. Due tomorrow btw.

The question is:

Suppose that is continuous everywhere, exists for , and exists. Prove that exists.
Professor said to use L'Hôpital's rule if we get stuck.

I started by trying to let and working through an argument using L'Hôpital's rule, but I made some extra assumptions and such which dont seem to be reasonable. Any nudges in the right direction would be great. My work is pretty useless...and ultimately it seems like I wouldnt be proving the right thing anyway. I feel like I am overthinking (as usual) what would be an otherwise simple and elegant proof. If anyone really wants to see my attempt, I will post it, but it is almost certainly incorrect. Thanks in advance.


edit: I know intuitively that it must exist, since is continuous and exists everywhere but (for sure). And since the two-sided limit exists at , either is continuous at or has a removable discontinuity there, but it must be defined there because is continuous at and does not have a corner there. There cannot be a corner at because the two-sided limit exists, so we know that the derivative must exist there (as is continuous without a corner)... but I do not see how L'Hôpital would come into play in a formal proof?

Originally Posted by xifentoozlerix

Suppose that is continuous everywhere, exists for , and exists. Prove that exists.

The first step that I like to do a lot of times is to formally state the problem. I find that it helps:
"Let be continous and let be differenciable on (possibly at also, but we do not know this). Suppose also that exists. Prove that is differenciable at zero".

We want to show " is differenciable at zero", which means we want to show that the limit exists.

Note the numerator is . This is where we are using the countinuity condition. The denominator is zero as well. Thus, we can use L'Hopital here because the numerator is differenciable. Thus, . And this exists by the stated problem.

Thank you so much....after sleeping on it I had a similar idea, but you make it seem so simple now (it always seems that way though...simple when you know how).