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" :)
Answer
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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