# Mysterious limit of a function

this is driving me crazy. I need to solve this:

Here's the riddle: Since it's (0/0), I do L'Hôpital's rule, which means I get to:

But the limit of this = 0 (after one more use of L'Hôpital's rule). And that is not the correct answer.

HOWEVER, if I just do the derivative of the integral, but LEAVE the denominator as is, then I get this:

And after some more use of L'Hôpital's rule, this actually comes out to be (-5) - which is supposed to be the correct answer.

So I don't understand why when I use L'Hôpital's rule on the numerator alone - it works, but if I use it on both numerator and denominator (which is how you're supposed to..) - it doesn't.

Would appreciate the solution of this "mystery" :)

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By the way, there is a chance that the "correct" solution given by our teacher is a mistake.. but I didn't want to rely on that unless I get some backup :) – Cauthon Apr 1 '14 at 9:57
I'll back you up --- I think 0 is correct. – Gerry Myerson Apr 1 '14 at 9:58
I think that $-5$ is correct(using Taylor). – Claude Leibovici Apr 1 '14 at 10:01
OK I gather from most replies that indeed the solution given is incorrect. I was confused because with a slight change it was correct (hence the origin of the mistake probably). Thanks everyone. – Cauthon Apr 1 '14 at 10:05
I confess I made a stupid mistake ! – Claude Leibovici Apr 1 '14 at 10:41

this actually comes out to be (-5) - which is supposed to be the correct answer.

It isn't.

If we Taylor-expand the integrand, we get

$$\int_0^x \frac{e^{-5t^2}-1}{t}\,dt = \int_0^x \frac{-5t^2 + O(t^4)}{t}\,dt = \int_0^x -5t + O(t^3)\,dt = -\frac{5}{2}x^2 + O(x^4).$$

The denominator is

$$\sqrt{1+2x}-1 = \left(1+x - \frac{x^2}{2} + O(x^3)\right) - 1 = x + O(x^2),$$

so the limit is indeed $0$.

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Great, thanks, I forgot to try using Taylor and compare. – Cauthon Apr 1 '14 at 10:07

You are right... We have to differentiate both numerator and denominator...your teacher did a silly mistake...

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