Calculate the limit (verifying my answer). I have to calculate
$$\lim_{x\rightarrow 1^+} \frac{\sin(x^3-1)\cos(\frac{1}{1-x})}{\sqrt{x-1}}$$
Making the substitution $x-1=y$ and doing the math, I get that,
$$=\lim_{y\rightarrow 0^+} \frac{\sin(y(y^2+3y+3))}{y(y^2+3y+3)}\cdot\cos\Big(\dfrac{1}{y}\Big)\cdot\sqrt{y}(y^2+3y+3)$$
Since the first fraction goes to $0$. I have to worry with the $\cos(1/y)$, but I realized that $\cos(x)$ is bounded above and below, then, this kind of function times something that goes to $1$ results in $0$ (I studied this theorem). Since $\sqrt{y}$ goes to $0$. Then, the asked limit is $0$. Is that correct?

hint
$$x^3-1=(x-1)(x^2+x+1)$$
hence
$$\sin(x^3-1)\sim 3(x-1) \;\;(x\to 1^+)$$
and
$$\frac{\sin(x^3-1)}{\sqrt{x-1}}\cos(\frac{1}{x-1} )\sim 3\sqrt{x-1}\cos(\frac{1}{x-1}) \;\;(x\to 1^+)$$
but
$$|\sqrt{x-1}\cos(\frac{1}{x-1})|\le \sqrt{x-1}$$
thus the limit is zero.
