# Change of domain of a function to change variable in the limit

I am trying to prove the following statement: $$\lim_{x\to 1} \frac{\sin(x^2-1)}{x^2-1} = \lim_{h\to 0} \frac{\sin(h)}{h}$$ It might seem obvious but I have trouble going from one variable to the other. Please note that I do not know what the derivative is therefore I cannot use it to solve this problem. Now, the way I tried to complete this problem is shown next.

Assume that $\lim_{h\to 0} \frac{\sin(h)}{h} = \lim_{h\to 0} f(h)$ exists and is equal to $\alpha$. Then I know that for each $\epsilon > 0$ there is $\delta > 0$ so that if $0 < \vert h \vert < \delta$ then $\vert f(h) - \alpha \vert < \epsilon$. In general it is assumed that the domain of the function are all of the $h$ for which $f(h)$ makes sense. I want to use locality of limit. Therefore, I want to change domain of function $f(h)$ from $\mathbb{R} \setminus\{0\}$ to $(-1 ; +\infty)\setminus\{0\}$. Then, for each $\epsilon > 0$, I just take $\delta ' = \mathrm{min} (1, \delta)$ so that if $0 < \vert h \vert < \delta '$ then the function is both defined and $\vert f(h) - \alpha \vert < \epsilon$.

Now, because of my new domain, I want to say that every $h$ can be written as $h = x^2 - 1$ by finding some $x$. To make $x$ unique, I require that $x \geq 0$. Then, for each $\epsilon > 0$ I can find $\delta' > 0$ so that if $0 < \vert x^2-1 \vert < \delta'$ then $\vert f(x^2-1) - \alpha \vert < \epsilon$.

Now, I notice that if $\vert x^2-1 \vert = \vert x-1 \vert \vert x+1 \vert < \delta'$ then it must be true that $\vert x-1 \vert < \frac{\delta '}{\vert x+1 \vert} \leq \delta '$. Therefore I can say that for each $\epsilon > 0$ I can find $\delta' > 0$ so that if $0 < \vert x-1 \vert < \delta'$ then $\vert f(x^2-1) - \alpha \vert < \epsilon$. But that is equivalent of saying that $\lim_{x\to 1} f(x^2-1) = \alpha$.

Therefore, I think I have proved what I wanted.

Questions : 1) Please can someone check my solution and justify all of the steps. 2) Maybe there is an easier solution? 3) How to deal with limits that tend $x$ to some $a$ but function is of the form $f(x^2)$ - by that I mean that there are only quadratic powers of $x$ in the values of the function. I am unsure about them because going to quadratic powers my domain changes and I do not know how to deal with it.

Thanks!

• Also I would like to receive feedback from people that downvoted my question so I can make it better next time. Commented Feb 18, 2017 at 12:02
• sadly, some people downvote without leaving feedback, but if you want mine, I think this is a good question. Well posed and full of effort :-) Commented Feb 18, 2017 at 12:05
• Jeah, good question! There is really no reason to downvote it (+1).
– user339727
Commented Feb 18, 2017 at 12:24
• This problem of downvotes for good questions is just awful. $\to (+1)$. Commented Feb 18, 2017 at 12:56
• +1 for the way you have presented your efforts. You should have a look at the general theorem concerning change of variables for limit evaluation mentioned in this answer math.stackexchange.com/a/1073047/72031 (and try to prove the theorem yourself). Commented Feb 18, 2017 at 19:07

A way I prefer (specific to this problem, though the general technique is a good one) because it doesn't require any substitutions which might be dodgy: note that $x^2-1 = (x+1)(x-1)$, so your left-hand limit is equal to $$\frac{1}{2} \lim_{x \to 1} \frac{\sin(x^2-1)}{x-1}$$ By the substitution $x-1 = u$, this is $$\frac{1}{2} \lim_{u \to 0} \frac{\sin[(u+2)u]}{u}$$
This is half the derivative of the function $x \mapsto \sin[(x+2)x]$ at $x=0$; that derivative is (by the chain rule) $(2x+2) \sin'[(x+2)x]$ evaluated at $x=0$, which is precisely $$2 \sin'(0) = 2 \lim_{h \to 0} \frac{\sin h}{h}$$ as required. (At no point did we need to calculate the value of this limit, of course, but one could extract it easily from the Taylor-series fact that $\sin' = \cos$.)