# Math Help - limit

1. ## limit

What is the limit of x->1 sinpi*x/(1-x)^2?
I want to solve it without using L'Hospital rule.

2. ## Re: limit

Hint: use the identity $\sin(\pi-\theta)=\sin(\theta)$ and $\lim_{x\to c}(f(x)\cdot g(x))=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)$ and a two-sided limit. What do you find?

3. ## Re: limit

sin(pi-x)*x*(pi-x)/(1-X)(1+x)(pi-x)=x*(pi-x)/(1-x)(1+x)?

4. ## Re: limit

No...I'm sorry but that makes no sense to me...please read my post above again and apply the identities I gave...

5. ## Re: limit

which one is f(x) or g(x)?

7. ## Re: limit

We are given to evaluate:

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

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

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

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

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

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

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

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

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

Applying (1), we are left with:

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

Now, do we have:

$\lim_{u\to0^{-}}\frac{1}{u}=\lim_{u\to0^{+}}\frac{1}{u}$ ?

8. ## Re: 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.

9. ## Re: limit

You have incorrectly applied the trigonometric identity.

10. ## Re: limit

-sin(pi*(x-1))
= --------------
(1-x)^2

sin(pi*(1-x))
= ------------- ( since -sin(y) = sin(-y) )
(1-x)^2

11. ## Re: 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.

12. ## Re: limit

I couldn't get it.I am sorry.

13. ## Re: 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...

14. ## Re: limit

edit: web error...sorry.