What is the limit of the given fraction in Calculus 1 homework?

[youngstudent16]

Homework Statement

The limit
$\lim_{x\to\pi}\frac{xcos\frac{x}{2}}{\pi^{2}-x^{2}}$
Can be expressed as a fraction. Solve

2. Relevant equation

3. The Attempt at a Solution

EDIT
See new post for solution

 

[laplacean]

Keep in mind, when taking the derivative, that cos(pi/2) is a constant.

 

[youngstudent16]

Keep in mind, when taking the derivative, that cos(pi/2) is a constant.

I know I'm getting 0 as the final answer that's not right though it should be a fraction of two relatively prime positive integers.

 

[youngstudent16]

I know I'm getting 0 as the final answer that's not right though it should be a fraction of two relatively prime positive integers.

My mistake I typed the question wrong

 

[laplacean]

The derivative of the top, when derived correctly, will yield a prime number. Just derive xcos(pi/x) as you would derive an x with a constant coefficient. For the bottom, pi^2 is also a constant and should go to 0 when derived. All you have left to deal with would be the -x^2.

 

[youngstudent16]

The derivative of the top, when derived correctly, will yield a prime number. Just derive xcos(pi/x) as you would derive an x with a constant coefficient. For the bottom, pi^2 is also a constant and should go to 0 when derived. All you have left to deal with would be the -x^2.

sorry it should be cos(x/2) I read it wrong

 

[Student100]

I know I'm getting 0 as the final answer that's not right though it should be a fraction of two relatively prime positive integers.

Why is zero incorrect?

 

[Student100]

sorry it should be cos(x/2) I read it wrong

Then that makes more sense, what do you get now?

 

[youngstudent16]

Why is zero incorrect?

I'm sorry I typed it wrong trying again with Let t=x−π. Then x=t+π,

 

[laplacean]

I did not get zero as an answer. A place where you could be making a computational mistake would be in the derivation of xcos(x/2).

 

[Student100]

I'm sorry I typed it wrong trying again with Let t=x−π. Then x=t+π,

Can you go ahead and retype the whole thing? The original limit was zero, the second correction gave a finite value, but retype everything.

 

[laplacean]

That would be much appreciated (:

 

[youngstudent16]

Then that makes more sense, what do you get now?

Let $t=x-\pi$ Then $x=t+\pi$

$\lim_{x\to \pi}\frac{xcos\frac{x}{2}}{\pi^{2}-x{^2}}$$=\lim_{t\to 0}\frac{(\pi+t)cos\frac{(\pi+t)}{2}}{\pi^{2}-(\pi+t){^2}}$$=\lim_{t\to 0}\frac{(\pi+t)sin\frac{t}{2}}{t^{2}+2\pi t}$

$=\lim_{t\to 0}\left(\frac{\pi + t}{2(t+2\pi)}\cdot \frac{sin\frac{t}{2}}{\frac{t}{2}}\right)$$=\frac{1}{4}$Latex thing is neat will use in future

 

[Dopplershift]

Can't you use L'hospital's rule since you have an indeterminate limit?

Where lim _x→π = f(x)/g(x) is also = lim _x→π f'(x)/g'(x)

So just take the derivative of the top divided by the derivative of the bottom. Don't for get chain rule and that π² is a constant, do not change π² into 2π. A lot of calculus students make that mistake.

Also remember the derivative of cosθ is -sinθ (A lot of calculus students forget the negative as well)

 

[Student100]

Can you? Sure. His solution is fine though.

$\lim_{x\to \pi}\frac{xcos\frac{x}{2}}{\pi^{2}-x{^2}}$

=

$\lim_{x\to \pi}\frac{xsin\frac{x}{2}}{4x}$

=

$\frac{1}{4}$