calculating limits using epsilon-delta definition

Question

$$\lim_{x \to -1} \left|\frac{2-x}{1+x}\right|=+∞$$ let $$f(x)=\left|\frac{2-x}{1+x}\right|$$ then $∀M∈ℝ>0, \exists\delta>0,\forall x\in D_f \left(\left|x+1 \right|<\delta\Longrightarrow \left|\frac{2- x}{1+x}\right|>M\right)$ take $\delta\le1$ implies:$$2<\left|x-2\right|<4$$$$M<\frac{\left|x- 2\right|}{\left|1+x\right|}<\frac{4}{\left|1+x\right|}$$$$\left|1+x\right|<\frac{4}{M}$$ hence $$\delta\le\min\left\{1,\frac{4}{M}\right\}$$ is it right? the other one is as follows: $$\lim_{x \to 0}\frac{\sin\left(x-1\right)}{x\left(x-1\right)}=∞$$ let $$g(x)= \frac{\sin\left(x-1\right)}{x\left(x-1\right)} $$ then $∀M∈ℝ>0, \exists\delta>0,\forall x\in D_g \left(\left|x+1 \right|<\delta\Longrightarrow |\frac{\sin\left(x-1\right)}{x\left(x-1\right)}\right|>M)$ but I don't know how to start, any hint or proof is appreciated.

Answer 1

For the first one, you've proved that $\frac{4}{|1+x|}>M$, which is not what you want. You need to use the lower bound for $|x-2|$. Since $|x-2|>2$, $\frac{|x-2|}{|x+1|}>\frac{2}{|x+1|}$ and you want $\frac{2}{|x+1|}>M$, so choose $\delta\leq \frac2M$. Do you see your mistake?

As for the second one, you need the absolute value of the expression to make the limit true; otherwise it diverges to $\pm \infty$ from both sides of $0$. Next, write $\frac{|\sin(x-1)|}{|x-1|}=\frac{|\sin(1-x)|}{|1-x|}$, and take $\delta \leq \frac14$, so you have $|x|<\frac14 \Leftrightarrow -\frac14<x<\frac14$. Then use that $\sin t$ is increasing on $[0,\pi/2]$. It follows that $\sin(1-x)\in (\sin\frac{3}{4}, \sin\frac54)$ and $1-x\in(\frac34,\frac54)$, and note that $\sin\frac34 >\sin\frac{\pi}6=\frac12$. Can you continue now?