Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

enter image description here

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

enter image description here

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:

enter image description here

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" :)

share|improve this question
1  
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 :) –  Crumbs Apr 1 at 9:57
    
I'll back you up --- I think 0 is correct. –  Gerry Myerson Apr 1 at 9:58
    
I think that $-5$ is correct(using Taylor). –  Claude Leibovici Apr 1 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. –  Crumbs Apr 1 at 10:05
    
I confess I made a stupid mistake ! –  Claude Leibovici Apr 1 at 10:41

2 Answers 2

up vote 2 down vote accepted

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 $0$.

share|improve this answer
    
Great, thanks, I forgot to try using Taylor and compare. –  Crumbs Apr 1 at 10:07

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.