I couldn't find this particular solution on here; I apologise in advance if it has been posted before.
I know that $\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$
I am asked to find this limit, using the limit above and using a change of variable, so not using L'Hôpital's rule: $$\lim_{x \rightarrow 1} \frac{\sin(x^2 -1)}{x-1}$$
I've tried a few things, such as:
$$\lim_{x \rightarrow 1} \frac{\sin(x^2 -1)}{x-1} = \lim_{x \rightarrow 1} \frac{\sin((x-1)(x+1))}{x-1}$$
From the first expression, it is evident that as x is approaching 1, the nominator is approaching 0, and so is the denominator. Let $u = x – 1$.
Then: $$\lim_{u \rightarrow 0} \frac{\sin(u(u+1)}{u} = \lim_{u \rightarrow 0} \frac{\sin(u^2 + u)}{u}$$
But this doesn't seem to get me any further from where I started… Could someone please give me a hint? Thank you.
Best Answer
Try $${\sin (x^2-1) \over x-1} = {(x+1)\sin (x^2-1) \over (x+1)(x-1)} = (x+1){\sin (x^2-1) \over x^2-1} $$