{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
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