# Homework Help: {Calculus 1} Find the limit

1. Aug 9, 2015

### youngstudent16

1. The problem statement, all variables and given/known data
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

Last edited: Aug 10, 2015
2. Aug 9, 2015

### laplacean

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

3. Aug 9, 2015

### youngstudent16

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

4. Aug 9, 2015

### youngstudent16

My mistake I typed the question wrong

5. Aug 9, 2015

### 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.

6. Aug 9, 2015

### youngstudent16

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

7. Aug 9, 2015

### Student100

Why is zero incorrect?

8. Aug 9, 2015

### Student100

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

9. Aug 9, 2015

### youngstudent16

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

10. Aug 9, 2015

### 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).

11. Aug 10, 2015

### Student100

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

12. Aug 10, 2015

### laplacean

That would be much appreciated (:

13. Aug 10, 2015

### youngstudent16

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

Last edited: Aug 10, 2015
14. Aug 10, 2015

### 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)

15. Aug 10, 2015

### Student100

Can you? Sure. His solution is fine though.

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

=

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

=

$\frac{1}{4}$

Last edited: Aug 10, 2015