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?
 A: 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.
A: First of all, since $ \left(\forall x\in\left]1,+\infty\right[\right),\ \left|\cos{\left(\frac{1}{x-1}\right)}\right|\leq 1 $, we get that $ \left(\forall x\in\left]1,+\infty\right[\right),\left|\sqrt{x-1}\cos{\left(\frac{1}{x-1}\right)}\right|\leq\sqrt{x-1} $, meaning : $ \lim\limits_{x\to 1^{+}}{\sqrt{x-1}\cos{\left(\frac{1}{x-1}\right)}}=0 \cdot $
Thus,
\begin{aligned} \lim_{x\to 1^{+}}{\frac{\sin{\left(x^{3}-1\right)}\cos{\left(\frac{1}{x-1}\right)}}{\sqrt{x-1}}}&=\lim_{x\to 1^{+}}{\frac{\sin{\left(x^{3}-1\right)}}{x^{3}-1}\left(x^{2}+x+1\right)\sqrt{x-1}\cos{\left(\frac{1}{x-1}\right)}}\\ &=1\times 3\times 0 \\ \lim_{x\to 1^{+}}{\frac{\sin{\left(x^{3}-1\right)}\cos{\left(\frac{1}{x-1}\right)}}{\sqrt{x-1}}}&=0\end{aligned}
A: $(x^3-1)=(x-1)(x^2+x+1)$;
$|\dfrac{\sqrt{x-1}(x^2+x+1)\sin (x^3-1)}{(x^3-1)}\cdot$
$\cos (\frac{1}{1-x})|=$
$(\sqrt{x-1}(x^2+x+1))\cdot \dfrac{\sin(x^3-1)}{(x^3-1)}$
$\cdot |\cos (\frac{1}{1-x})|.$
Take the limit $x \rightarrow 1^+$.
Note :
1) Limit of first term $=0$;
2) Use $\lim_{y \rightarrow 0^+}\dfrac{\sin y}{y}=1$;
3)$ |\cos z| \le 1$ for $z$ real.
