$$\begin{split} \lim_{x\to 1}\frac{\sin(x^2-1)}{x-1} &= \lim_{x\to 1}\frac{(x+1)\sin(x^2-1)}{(x^2-1)} \\ & = 2\lim_{x\to 1}\frac{\sin(x^2-1)}{x^2-1} \\ & = 2\lim_{h\to 0}\frac{\sin(h)}{h} \\ & = 2 \end{split}$$
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. $
The main idea of limits is that, one wants to figure out an undefined part of function. But the thing about limits is that not all of them can be evaluated just by looking at function itself. The idea is to evaluate this limit by converting the limit into various smaller limits whose evaluated values are know to us. The x^2 -1 can itself be regarded as a another variable h. And that gives you a limit that one already knows to evaluate.
