What is the limit of x->1 sinpi*x/(1-x)^2?

I want to solve it without using L'Hospital rule.

Printable View

- Apr 26th 2013, 12:41 AMkastamonulimit
What is the limit of x->1 sinpi*x/(1-x)^2?

I want to solve it without using L'Hospital rule. - Apr 26th 2013, 12:48 AMMarkFLRe: limit
Hint: use the identity $\displaystyle \sin(\pi-\theta)=\sin(\theta)$ and $\displaystyle \lim_{x\to c}(f(x)\cdot g(x))=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)$ and a two-sided limit. What do you find?

- Apr 26th 2013, 01:45 AMkastamonuRe: limit
sin(pi-x)*x*(pi-x)/(1-X)(1+x)(pi-x)=x*(pi-x)/(1-x)(1+x)?

- Apr 26th 2013, 01:52 AMMarkFLRe: limit
No...I'm sorry but that makes no sense to me...please read my post above again and apply the identities I gave...

- Apr 26th 2013, 04:13 AMkastamonuRe: limit
which one is f(x) or g(x)?

- Apr 27th 2013, 12:46 AMkastamonuRe: limit
What is the answer then?

- Apr 27th 2013, 01:51 AMMarkFLRe: limit
We are given to evaluate:

$\displaystyle \lim_{x\to1}\frac{\sin(\pi x)}{(1-x)^2}$

The first thing I am looking for is a way to apply the well-known result:

(1) $\displaystyle \lim_{u\to0}\frac{\sin(u)}{u}=1$

Seeing the denominator has $\displaystyle 1-x$ as a factor, this takes me to $\displaystyle \sin(\pi x)=\sin(\pi-\pi x)=sin(\pi(1-x))$. So we now have:

$\displaystyle \lim_{x\to1}\frac{\sin(\pi(1-x))}{(1-x)^2}$

Now, let's use the substitution $\displaystyle x=1-x$ and we have:

$\displaystyle \lim_{u\to0}\frac{\sin(\pi u)}{u^2}$

Now let's finish rewriting the expression so we can apply (1):

$\displaystyle \lim_{u\to0}\frac{\pi\sin(\pi u)}{\pi u^2}=\lim_{u\to0}\frac{\sin(\pi u)}{\pi u}\cdot\frac{\pi}{u}=\lim_{u\to0}\frac{\sin(\pi u)}{\pi u}\cdot\lim_{u\to0}\frac{\pi}{u}$

Applying (1), we are left with:

$\displaystyle \pi\lim_{u\to0}\frac{1}{u}$

Now, do we have:

$\displaystyle \lim_{u\to0^{-}}\frac{1}{u}=\lim_{u\to0^{+}}\frac{1}{u}$ ? - Apr 27th 2013, 09:40 AMkastamonuRe: limit
sin(pi*x) sin(pi*(x-1)+pi)

--------- = ----------------

(1-x)^2 (1-x)^2

we can replace (1-x) with u.

I am stuck in here. - Apr 27th 2013, 10:05 AMMarkFLRe: limit
You have incorrectly applied the trigonometric identity.

- Apr 27th 2013, 10:10 PMkastamonuRe: limit
Yes I found the answer.

-sin(pi*(x-1))

= --------------

(1-x)^2

sin(pi*(1-x))

= ------------- ( since -sin(y) = sin(-y) )

(1-x)^2 - Apr 27th 2013, 10:18 PMMarkFLRe: limit
Okay, but that's really unnecessary...did you actually read my post above where I showed you what to do except for the trivial last step? I hate to think my efforts are in vain. :D

- Apr 27th 2013, 10:27 PMkastamonuRe: limit
I couldn't get it.I am sorry.

- Apr 27th 2013, 10:33 PMMarkFLRe: limit
What exactly about what I posted don't you get...I want to help, but my main objective is for you to truly understand...tell me where I lose you in my post...

- Apr 27th 2013, 10:44 PMMarkFLRe: limit
edit: web error...sorry.

- Apr 27th 2013, 11:27 PMkastamonuRe: limit
answer is pi/2. How can I get this answer by 1/u?