Confused about a step in trying to show $\lim_{x\to 1}\frac{\sin(x^2-1)}{x-1} = 2$ $$\begin{split}
\lim_{x\to 1}\frac{\sin(x^2-1)}{x-1} &= \lim_{x\to 1}\frac{(x+1)\sin(x^2-1)}{(x^2-1)} \\
& = 2\lim_{x\to 1}\frac{\sin(x^2-1)}{x^2-1} \\
& = 2\lim_{h\to 0}\frac{\sin(h)}{h} \\
& = 2
\end{split}$$
I'm confused about how $\lim_{x\to 1}\frac{\sin(x^2-1)}{x^2-1} = \lim_{h\to 0}\frac{\sin(h)}{h}$.  What's the reasoning behind it and what is the more general rule for this?  I'm a fool trying to learn how limits work rigorously - any help would be appreciated.

There is a basic and useful result, usually not well explained, that states: If
i) $I,J$ are open intervals containing $a,b$ respectfully,
ii) $f$ is defined in $J\setminus \{b\}$ and
iii) $g:I\to J$ is $1–1$ and continuous, with $g(a)=b,$
then
$$\lim_{y\to b}f(y)=L\, \iff \lim_{x\to a}f(g(x))=L.$$
In your problem, $f(y) = (\sin y)/y$ with $b=0,$ and $g(x)=x^2-1$ with $a=1. $
