# L'Hopital's Rule 0/0 with an integral in the numerator

• March 24th 2008, 05:05 PM
paulrb
L'Hopital's Rule 0/0 with an integral in the numerator
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.
• March 24th 2008, 05:41 PM
topsquark
Quote:

Originally Posted by paulrb
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.

Use the Fundamental Theorem of Integral Calculus to take the derivative of your numerator:
$\frac{d}{dx} \int_0^x \sqrt{1 + sin(t)}~dt = \sqrt{1 + sin(x)}$

-Dan
• March 25th 2008, 06:33 AM
colby2152
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.
• March 25th 2008, 07:00 AM
topsquark
Quote:

Originally Posted by colby2152
Following what Dan provided for you... what happens with the numerator as x approaches zero? Sine goes to one, and so the numerator goes to the square root of two. The "bottom" tends to get smaller and smaller while the "top" approaches a constant value, so the value of this function goes to infinity as x approaches zero.

I think there is a mistake here. When doing the L'Hopital's step we get
$\lim_{x \to 0} \frac{\int_0^x \sqrt{1 + sin(t)}~dt}{x} \to \frac{0}{0}$

So
$\lim_{x \to 0} \frac{\int_0^x \sqrt{1 + sin(t)}~dt}{x}$

$= \lim_{x \to 0} \frac{ \frac{d}{dx} \int_0^x \sqrt{1 + sin(t)}~dt}{\frac{d}{dx}x}$

$= \lim_{x \to 0} \frac{\sqrt{1 + sin(x)}}{1} = 1$

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?

-Dan
• March 25th 2008, 07:14 AM
colby2152
Quote:

Originally Posted by colby2152
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" tends to get smaller and smaller while the "top" approaches a constant value, so the value of this function goes to one as x approaches zero.

Whoops, I forgot to take the derivative of the "bottom"! Thanks Dan. I also evaluated sine incorrectly at zero.

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.
• March 25th 2008, 09:12 AM
topsquark
Quote:

Originally Posted by colby2152
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.

Well I haven't really done out the thing, but as the first term in the expansion is 0, all the rest of the terms are divisible by x. etc. It ought to work, but I was really hoping that someone would have a "better" way to do it.

-Dan
• March 25th 2008, 10:14 AM
Soroban
Hello, paulrb!

There seems to be a gap in your education . . .

Quote:

$\lim_{x\to0}\,\frac{\int^x_0\sqrt{1+\sin t}\,dt}{x}$
We're expected to know that: . $\frac{d}{dx}\left[\int^x_0 f(t)\,dt\right] \;=\;f(x)$

Hence: . $\frac{d}{dx}\left[\int^x_0\sqrt{1 +\sin t}\:dt\right] \;=\;\sqrt{1+\sin x}$

Can you finish it now?

• March 25th 2008, 02:07 PM
paulrb
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?
• March 25th 2008, 02:35 PM
topsquark
Quote:

Originally Posted by paulrb
Just curious, why would someone think L'Hopital's rule should only be used as a last resort?

Honestly I don't think there is any reason to be bashful about using L'Hopital's rule. But some of the members here, I think, are so good at doing the limits they find that doing the limit without L'Hopital is much more of a challenge and thus it is better to do it that way. (Kind of like using a hammer to crack open a nut: it lacks subtlety.) Point in fact, you do learn a lot of Math by avoiding L'Hopital's rule. I have anyway. But if you are just trying to do your homework, use L'Hopital.

-Dan