Calculate the limit (verifying my answer).

Question

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?

Answer 1

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}

Answer 2

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.

Answer 3

$(x^3-1)=(x-1)(x^2+x+1)$;

$|\dfrac{\sqrt{x-1}(x^2+x+1)\sin (x^3-1)}{(x^3-1)}\cdot$

$\cos (\frac{1}{1-x})|=$

$(\sqrt{x-1}(x^2+x+1))\cdot \dfrac{\sin(x^3-1)}{(x^3-1)}$

$\cdot |\cos (\frac{1}{1-x})|.$

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.