$\displaystyle \lim_{x\to1}\frac{sin(x^2-1)}{x-1}$
I tried making t=x-1
but I was left with sin(t(x+1)) which I didn't know what to do with
t = x - 1 we get $\displaystyle \lim_{t\to 0} \frac{\sin t(t+2)}{t}$.
We get, $\displaystyle \lim_{t\to 0}\frac{\sin t^2 \cos 2t + \cos t^2\sin 2t}{t}$
Replace $\displaystyle \cos 2t $ and $\displaystyle \cos t^2$ by 1 because that is their limit and we have:
$\displaystyle \lim_{t\to 0}\frac{\sin t^2}{t} + \frac{\sin 2t}{t} = \lim_{t\to 0}\frac{\sin t^2}{t^2}\cdot t + \frac{\sin 2t}{2t} \cdot 2 = 2$
There are many ways to approach these limits.
You could let $\displaystyle t=x^{2}-1$
$\displaystyle x=\sqrt{t+1}$
$\displaystyle \lim_{t\rightarrow{0}}\frac{sin(t)}{\sqrt{t+1}-1}$
Now, multiply by the conjugate:
$\displaystyle \lim_{t\rightarrow{0}}\frac{\sqrt{t+1}sin(t)}{t}+\ lim_{t\rightarrow{0}}\frac{sin(t)}{t}$
We can see the familiar sin(t)/t limit. Which equals 1.
So you have the limit 1+1=2
Here's another attempt.
$\displaystyle \lim_{x \to 1} \frac{{\sin \left( {x^2 - 1} \right)}}
{{x - 1}} = \lim_{x \to 1} \frac{{\sin \left( {x^2 - 1} \right)}}
{{x^2 - 1}} \cdot (x + 1).$
You can see the limit of the first factor exists and it's equal to 1, just make $\displaystyle u=x^2-1$ to check it.
From here you can easily see that the requested value of the limit is 2.