Limit with sinus function How to compute these two limits $$\lim_{x \to 1}\sin\frac{1}{x-1}$$ and $$\lim_{x \to 1}(x-1) \sin \frac{1}{x-1}?$$ Squeeze theorem should do the thing but I do not know how?
 A: 1)
$\lim \limits_{x \to 1} \sin\left(\frac{1}{x-1}\right)=\lim \limits_{x \to 0} \sin\left(\frac{1}{x}\right)=\lim\limits_{x \to \infty} \sin(x)$
Which does not exist! (Its up to you to take two subsequences which do converge to different limit-points)
2)
$\lim \limits_{x \to 1} \left| (x-1)\sin\left(\frac{1}{x-1}\right)\right|=\lim \limits_{x \to 1} | (x-1) |\left|\sin\left(\frac{1}{x-1}\right)\right|\leq   \lim \limits_{x \to 1} | (x-1) |=0$
A: The first limit does not exist! Make substitution $u=\frac{1}{x-1}$, so when $x\to1\Rightarrow u\to\infty$ and:
$$\lim_{u\to \infty}\sin\left(u\right)=?$$
(It is not defined because the sine oscillates).
For the second limit,
$$\lim_{x\to 1}(x-1)\sin\left(\frac{1}{x-1}\right)$$
Make the substitution $u=\frac{1}{x-1}$, so when $x\to1\Rightarrow u\to\infty$ and:
$$\lim_{u\to \infty}\frac{\sin\left(u\right)}{u}$$
But $|\sin(u)|<1$, so
$$\lim_{u\to\infty}\frac{-1}{u}\leq\lim_{u\to \infty}\frac{\sin\left(u\right)}{u}\leq\lim_{u\to\infty}\frac{1}{u}$$
By the squeeze theorem, both leftmost and rightmost limits tend to zero, so
$$0\leq\lim_{u\to \infty}\frac{\sin\left(u\right)}{u}\leq0\Rightarrow\lim_{u\to \infty}\frac{\sin\left(u\right)}{u}=0$$
