Appropriate change of variable for limit (sine function)

Question

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.

Answer 1

Try $${\sin (x^2-1) \over x-1} = {(x+1)\sin (x^2-1) \over (x+1)(x-1)} = (x+1){\sin (x^2-1) \over x^2-1} $$

Answer 2

A bit of context:

We show that $\lim_{x \rightarrow 1} (x+1)\dfrac{\sin(x^2-1)}{x^2-1}$ exists.

Suffices to show that the limit of each factor of the above product exists.

First factor $(x+1)$ is no problem.

Need to show:

1) For $\epsilon >0$ given, there is a $\delta >0$ such that

$|x-1|<\delta$ implies $|\dfrac{\sin (x^2-1)}{x^2-1}|<\epsilon$.

We know :

2) $\lim_{y \rightarrow 0} \dfrac{\sin y}{y} =1$, i.e.

For $\epsilon >0$ there is a $\Delta$ such that

$|y|<\Delta$ implies $|\dfrac{\sin y}{y} -1|< \epsilon.$

Set $y:= x^2-1.$

3) $y=x^2- 1$ is continuos at $x=1$:

Let $\epsilon_1 >0$ be given, then there is a $\Delta_1$

such that

$|x-1| < \Delta_1$ implies $|x^2-1|<\epsilon_1$.

4) Now choose

$\epsilon_1 < \Delta$, and $\delta = \Delta_1.$

Then:

$|x-1|< \Delta_1= \delta$

implies $|x^2-1| <\epsilon_1 <\Delta $

Implies $|\dfrac{\sin (x^2-1)}{x^2-1}| <\epsilon$.