<s>[INST] 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?
 [/INST] 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}
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