# L'Hospital's rule

• Nov 19th 2009, 06:36 PM
Eternal
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?
• Nov 19th 2009, 06:40 PM
Alterah
If your function is: $x*sin\frac{\pi}{x}$ then this isn't in the form of 0/0 or infinity/infinity and therefore you can't use L'Hopital's rule.
• Nov 19th 2009, 06:42 PM
lvleph
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.
• Nov 19th 2009, 06:42 PM
Eternal
Quote:

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

Well, then can you explain why the answer is pi?
• Nov 19th 2009, 06:50 PM
Haven
But the function is of indeterminate form. I.e, $0*\infty$ so we can rewrite the limit so it is of the form $\frac{0}{0}$

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

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

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

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

$\lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi$
• Nov 19th 2009, 06:52 PM
Alterah
Haven, you beat me to it. Took me a second to realize that it was actually an indeterminate form.
• Nov 19th 2009, 06:53 PM
lvleph
Quote:

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

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

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

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

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

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

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.
• Nov 19th 2009, 07:00 PM
Eternal
Quote:

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.

Thanks man. Didn't know you had to transform it like that. I've just started learning this subject.
• Nov 19th 2009, 07:03 PM
lvleph
Quote:

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

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.
• Nov 19th 2009, 07:24 PM
Eternal
Another question I'm confused on:

lim x->0+ Sin(x)lnx

How would I solve this? ln(0) is undefined...
• Nov 19th 2009, 07:31 PM
lvleph
Make the substitution $\ln x = u$. Then we have $\lim_{u \to -\infty}\sin (e^{u}) u$.
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# L'Hospital's rule

• Nov 19th 2009, 06:36 PM
Eternal
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?
• Nov 19th 2009, 06:40 PM
Alterah
If your function is: $x*sin\frac{\pi}{x}$ then this isn't in the form of 0/0 or infinity/infinity and therefore you can't use L'Hopital's rule.
• Nov 19th 2009, 06:42 PM
lvleph
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.
• Nov 19th 2009, 06:42 PM
Eternal
Quote:

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

Well, then can you explain why the answer is pi?
• Nov 19th 2009, 06:50 PM
Haven
But the function is of indeterminate form. I.e, $0*\infty$ so we can rewrite the limit so it is of the form $\frac{0}{0}$

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

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

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

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

$\lim_{x\rightarrow\infty} \pi\cos{\frac{\pi}{x}} = 1*\pi$
• Nov 19th 2009, 06:52 PM
Alterah
Haven, you beat me to it. Took me a second to realize that it was actually an indeterminate form.
• Nov 19th 2009, 06:53 PM
lvleph
Quote:

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

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

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

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

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

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

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.
• Nov 19th 2009, 07:00 PM
Eternal
Quote:

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.

Thanks man. Didn't know you had to transform it like that. I've just started learning this subject.
• Nov 19th 2009, 07:03 PM
lvleph
Quote:

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

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.
• Nov 19th 2009, 07:24 PM
Eternal
Another question I'm confused on:

lim x->0+ Sin(x)lnx

How would I solve this? ln(0) is undefined...
• Nov 19th 2009, 07:31 PM
lvleph
Make the substitution $\ln x = u$. Then we have $\lim_{u \to -\infty}\sin (e^{u}) u$.
user
The text passage contains a discussion about a mathematical problem and its solution, including a question and an answer provided by other users.

TYPE: SINGLE_QUESTION_WITH_ANSWER