Simple limit problem: $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$ While trying to help my sister with her homework she gave me the next limit: $$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$$
I know the conventional way of solving it would be (That's what i showed her):
$$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\lim_{x\to2}\left(\frac{x+2-4}{x^2-4}\right)=\lim_{x\to2}\left(\frac{x-2}{(x+2)(x-2)}\right)=\lim_{x\to2}\left(\frac{1}{x+2}\right)=\frac14$$
But she gave me the next answer:
$$\begin{align}
\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})&=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\lim_{x\to2}\frac{1}{x+2}\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\frac14\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-\lim_{x\to2}\frac{1}{x-2}\\
&=0
\end{align}$$
I actually couldn't explain her why is she wrong. Cause technically it looks fine. What am i missing? 

$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}$ -- it is true only if these limits exist.
