Two Undefined Limits?

Combine the three fractions together into one fraction:

$\large\displaystyle\lim_{x\to2}\left(\dfrac{3x-6}{x^2-4}\right)$ $\large =\displaystyle\lim_{x\to2}\left(\dfrac{3\cancel{(x-2)}}{(x+2)\cancel{(x-2)}}\right)$ $\huge =\dfrac{3}{4}=\boxed{0.75}$

In general for computing limit $\begin{aligned} \lim_{x \to 2}\left(\frac{2}{x-2}+\frac{1}{x+2}-\frac{8}{x^2-4}\right) &= \lim_{x \to 2}\left(\frac{2}{(2)-2}+\frac{1}{(2)+2}-\frac{8}{(2)^2-4}\right) \\&= 0.75 \space\space\ \square \end{aligned}$

$\large \text{FIN!!!}$

It is easy to see that $\dfrac{8}{(x+2)(x-2)}$ can be written as $\dfrac{2}{x-2}\ - \dfrac{2}{x+2}$ . Therefore, the terms with x-2 at the denominator cancel and we are left with $\large \lim_{x\to2} \dfrac{3}{x+2}$ , which is obviously $\dfrac{3}{4}$

Since $x$ approaches $2$ , substitute a value for $x$ that is very near to $2$ , let $x=1.9999$ . Substitute then use a calculator, so we have

$\dfrac{2}{1.9999-2}+\dfrac{1}{1.9999+2}-\dfrac{8}{1.9999^2-4} \approx \boxed{0.75}$