What is the limit of $f(x)=\frac{\sqrt{(x-2) \left(x^2+2 x-8\right)}}{x^2-4}$ when $x\to 0$?

Question

I tried to find out the value of limit but when I took $x_-\rightarrow0$ then a result is obtained again when I took $x_+\rightarrow0$ another result was obtained. Can anyone please explain why is this happening?

Is there anything wrong with the question or I was doing it wrong?

I am quite weak at determining limits

Edit: The problem stated above has been solved. The question was wrong instead of $x \to 0$ it should be $x \to 2$ (this will satisfy the question).

Answer 1

I'm sure you meant to evaluate the limit as $x \to 2$.

Factorize, and remember that $\sqrt {y^2}$ is not just $y$; it is $|y|$. This is crucial.

What's the behavior of the function $\frac {|y|}{y}$ near zero?

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Solution:

$$ \frac {\sqrt{(x-2)(x^2 +2x -8)}}{x^2 -4}=\frac {\sqrt{(x-2)(x+4)(x-2)}}{(x-2)(x+2)}= \frac {\sqrt {(x-2)^2} \sqrt {x+4} }{(x-2)(x+2)} = \frac {|x-2|}{x-2} \frac {\sqrt {x+4}}{x+2} $$
Now as $x \to 2$ the function $\frac {\sqrt {x+4}}{x+2} \to \frac {\sqrt {6}}{4}$, however the other half $\frac {|x-2|}{x-2}$ (aka $sign(x-2)$) is equal to $1$ as we approach $2$ from the right, and is equal to $-1$ as we approach $2$ from the left. Therefore the solution will be $-\frac {\sqrt {6}}{4}$ as $x \to 2^{\mathbf -}$ and $\frac {\sqrt {6}}{4}$ as $x \to 2^{\mathbf +}$.