This problem is just a collection of limits to solve in terms of $$\lim\limits_{x\rightarrow0}{\frac{\sin(x)}x}= \alpha$$
Now i solved all of them except for the last one, which is: $$\lim\limits_{x\rightarrow1}\left((x^2-1)^3\sin\left(\frac{1}{x-1}\right)^3\right)$$ I tried different factoring the expression but i just couldn't. So I went to the solutions manual and I just got more confused.
First of all, I think there's a mistake, because $$ \left|\sin\left(\frac{1}{x-1}\right)^3\right| \leq 1$$ is true even for $x = 0$, so i guess he meant for all $x \neq 1 $ But even then I don't quite get what he is doing. I graphed the function and indeed It seems like $x$ approaches $0$. However I just don't see why $ \left|\sin\left(\frac{1}{x-1}\right)^3\right| \leq 1$ would imply that. I wonder if it's anything like the $\epsilon ,\delta$ proofs that he gave before with $x\sin(1/x)$, so in this case, i would have to find some $\delta$ which guarantees $$|(x^2-1)^3| \leq \epsilon$$ if $0<|x-1|<\delta$
I didn't continue that idea because it seemed so long but if that's the only solution i guess i will have to do it. Is there any other way? Can you tell me why Spivak just gives that argument and calls it done? Thank you for advance.
Spivak is using this fact for $c=1$, $f(x)=(x^2-1)^3\to 0$ as $x\to 1$, and $g(x)=\sin\left(\frac{1}{x-1}\right)^3$ which is in fact bounded everywhere except at $x=1$: $|g(x)|\leq 1$. And you're right about the typo. It should be $x\ne 1$.
For a proof, you can use the squeeze theorem (you can also try an $\varepsilon$-$\delta$ proof). There is a neighborhood of $c$ and a positive number $M$ such that $|g(x)|\leq M$ for $x$ in that neighborhood. Then $|f(x)g(x)|\leq M|f(x)|$. Since $f(x)\to 0$, it follows by the squeeze theorem that $|f(x)g(x)|\to 0$ which implies that $f(x)g(x)\to 0$.