L'Hospital's rule

I'm having a difficulty differentiating this using L'Hospital's rule:

lim x->(infinite) xsin(pi/x)

Applying L'Hospital's rule and differentiating I got:

lim x->(infinite) xcos(pi/x) + sin(pi/x)

and that equals = (infinite)(1) + 0 = infinite

The correct answer in my textbook is just "pi"

Can someone tell me what I'm doing wrong?

If your function is: $\displaystyle x*sin\frac{\pi}{x}$ then this isn't in the form of 0/0 or infinity/infinity and therefore you can't use L'Hopital's rule.

Well, to use L'Hospital you must have an indefinite form. Second you differentiate each part seperately. Make sure you check what the indefinite forms are.

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Originally Posted by Alterah

If your function is: $\displaystyle x*sin\frac{\pi}{x}$ then this isn't in the form of 0/0 or infinity/infinity and therefore you can't use L'Hopital's rule.

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Well, then can you explain why the answer is pi?

But the function is of indeterminate form. I.e, $\displaystyle 0*\infty $ so we can rewrite the limit so it is of the form $\displaystyle \frac{0}{0} $

$\displaystyle x\sin{\frac{\pi}{x}} = \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}}$

we see $\displaystyle \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} = \frac{0}{0} $

apply L'Hospital's Rule
$\displaystyle \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} =
\lim_{x\rightarrow\infty} \frac{\frac{-\pi}{x^{2}}\cos{\frac{\pi}{x}}}{\frac{-1}{x^{2}}} = \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}}$

Since $\displaystyle \lim_{x\rightarrow\infty} \frac{1}{x} = 0 $

$\displaystyle \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi$

Haven, you beat me to it. Took me a second to realize that it was actually an indeterminate form.

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Originally Posted by Haven

But the function is of indeterminate form. I.e, $\displaystyle 0*\infty $ so we can rewrite the limit so it is of the form $\displaystyle \frac{0}{0} $

$\displaystyle x\sin{\frac{\pi}{x}} = \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}}$

we see $\displaystyle \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} = \frac{0}{0} $

apply L'Hospital's Rule
$\displaystyle \lim_{x\rightarrow\infty} \frac{\sin{\frac{\pi}{x}}}{\frac{1}{x}} =
\lim_{x\rightarrow\infty} \frac{\frac{-\pi}{x^{2}}\cos{\frac{\pi}{x}}}{\frac{-1}{x^{2}}} = \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}}$

Since $\displaystyle \lim_{x\rightarrow\infty} \frac{1}{x} = 0 $

$\displaystyle \lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi$

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The problem wasn't but could be transformed into indeterminate form. I was hoping the OP would see how to transform it, and then solve the problem.

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Originally Posted by lvleph

The problem wasn't but could be transformed into indeterminate form. I was hoping the OP would see how to transform it, and then solve the problem.

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Thanks man. Didn't know you had to transform it like that. I've just started learning this subject.

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Originally Posted by Eternal

Thanks man. Didn't know you had to transform it like that. I've just started learning this subject.

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Notice that the problem is exactly the same, but has been manipulated so that it looks different and you get one of the indeterminate forms.

Another question I'm confused on:

lim x->0+ Sin(x)lnx

How would I solve this? ln(0) is undefined...

Make the substitution $\displaystyle \ln x = u$. Then we have $\displaystyle \lim_{u \to -\infty}\sin (e^{u}) u$.