Find the limit as x approaches 0 of the integral of (1+sint)^(1/2)dt from 0 to x, divided by x. (x is in the denominator)
This is 0/0 format, since the integral in the numerator approaches 0 and the denominator approaches 0, but I don't understand how to get the answer. The integral is with respect to t, and I'm trying to take the derivative with respect to x. It doesn't make sense to me. Can anyone help? The answer in the back of the book says it's 1.
Following what Dan provided for you... what happens with the numerator as x approaches zero? Sine goes to zero, and so the numerator goes to one. The "bottom" is just one while the "top" approaches a constant value, so the value of this function goes to one as x approaches zero.
Just for the sake of curiosity, since there are so many on this forum that only use L'Hopital's rule as an utter last resort, is there a way to do this without using L'Hopital's rule? Like as in a MacLaurin series or something?
BTW, MacLaurin series will give you a polynomial plus a constant of one for the square root of one plus sine function. Problem with this is that the lone one from the MacLaurin series is divided by x which goes to zero resulting in an infinite result.
Thanks everyone. My problem was not remembering the Fundamental Theorem; I was thinking it only applied to variables of the same letter. However, upon reviewing it, I saw that the Fundamental Theorem uses x and t, not just x. After that it's easy.
Just curious, why would someone think L'Hopital's rule should only be used as a last resort?