# $\lim_{x \uparrow 1}f(x)$ and $\lim_{x \downarrow -1} f(x)$ of $\frac{\sqrt{(x-3)^2}-2}{1-x^2}, x \in D$

We look at the function $$f:D:= (-1,1) \subset \mathbb{R} \to \mathbb{R}$$ given by

$$f(x) := \frac{\sqrt{(x-3)^2}-2}{1-x^2}, x \in D$$

I want to find out $$\lim_{x \uparrow 1}f(x)$$ and $$\lim_{x \downarrow -1} f(x)$$

Expanding the nominator with $$(\sqrt{(x-3)^2}+2)$$ gives

$$= \frac{ (\sqrt{(x-3)^2}-2) (\sqrt{(x-3)^2}+2)}{ \sqrt{(x-3)^2}+2} = \frac{x^2-6x+5}{\sqrt{(x-3)^2}+2}$$

With the denominator we'd get

$$\lim_{x \uparrow 1} \frac{\frac{x^2-6x+5}{\sqrt{(x-3)^2}+2}}{1-x^2}$$

But then?

Since $$\vert x \vert = \sqrt{x^2}$$, we can simplify the expression $$\frac{\sqrt{(x-3)^2}-2}{1-x^2}$$:

$$\frac{\sqrt{(x-3)^2}-2}{1-x^2} = \frac{\vert x-3\vert-2}{1-x^2}$$

And since $$x \in (-1,1)$$, we can further simplify:

$$\frac{\vert x-3 \vert - 2}{1-x^2} = \frac{-(x-3) - 2}{1-x^2} = \frac{1-x}{1-x^2} = \frac{1-x}{(1-x)(1+x)}$$

Now if we take the limit:

$$\lim_{x \to 1} \frac{1-x}{(1-x)(1+x)} = \lim_{x \to 1} \frac{1}{1+x} = \frac{1}{2}$$

$$\lim_{x \to -1} \frac{1-x}{(1-x)(1+x)} = \lim_{x \to -1}\frac{1}{1+x} = \pm \infty$$ depending, if you go from the left or from the right side.

Hint: $$\sqrt{(x-3)^2}=|x-3|=3-x\,\,\text{for}\,\,x\in(-1,1)$$ Then you can simplify $$f(x)$$.