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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?

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2 Answers 2

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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.

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

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