Q:

How to compute $\lim_{x \to 1^{\pm}} \frac{4x \sqrt{\frac{(x^2-1)^2}{(x^2+1)^2}}}{x^4-1}$

How can I calculate in an easy way the following limit?  $$\lim_{x \to 1^{\pm}} \frac{4x \sqrt{\frac{(x^2-1)^2}{(x^2+1)^2}}}{x^4-1}$$

A:

Hint.  As Zach Gershkoff said, you may write, for $x\neq 1$:
$$
 \frac{4x \sqrt{\dfrac{(x^2-1)^2}{(x^2+1)^2}}}{x^4-1}= \frac{4x \dfrac{|x^2-1|}{x^2+1}}{x^4-1}=\frac{4x \dfrac{|x^2-1|}{x^2+1}}{(x^2-1)(x^2+1)}=\frac{|x^2-1|}{x^2-1}\frac{4x }{(x^2+1)^2}
$$ Hoping you can take it from here.