Solve these two limits $\lim_{x\to 1} \left(\frac{1}{x-1}-\frac{2}{x^2-1}\right)$ and $\lim_{x\to 0}\left(\frac{1}{x}-\frac{2}{|x|}\right)$

Question

Can someone help me to solve this limit please? it always lead to A/0 such that A is any number belong to R.

$$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{2}{x^2-1} \right)$$

also the left side part lead to same answer:

$$\lim_{x \to 0} \left( \frac{1}{x}-\frac{2}{|x|} \right)$$

Answer 1

Hint: Notice that $$\frac{1}{x-1} - \frac{2}{x^2-1} = \frac{1}{x-1} \left(1 - \frac{2}{x+1}\right) = \frac{1}{x-1} \left(\frac{x+1 - 2}{x+1}\right) = \frac{1}{x+1}$$

To the second one, just remember $$|x| = \begin{cases} x , \,\, \text{if}\,\, x\geq 0\\-x, \,\,\text{if}\,\,x<0 \end{cases}$$

and take the side limits.

Answer 2

Hint: $$\frac{1}{x-1}-\frac2{x^2-1}=\frac{x+1}{x^2-1}-\frac{2}{x^2-1}=\frac{x-1}{x^2-1}=\frac{(x-1)}{(x-1)(x+1)}=\cdots?$$

Answer 3

make a change of variable $x = 1 + h, h = x - 1.$ then $$\frac1{x-1}-\frac2{x^2-1} = \frac1h - \frac2{(1+h)^2 - 1} =\frac1h-\frac{2}{2h+h^2}=\frac{1}{(2+h)} \to \frac12$$ as $h \to 0.$