<s>[INST] 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? 
 [/INST] If $\lim_{x \to 2} \frac{1}{x-2}-\frac 4{x^2-4}=0$, that means $$\lim_{x \to 2} \frac{1}{x-2}- \lim_{x \to 2} \frac 4{x^2-4}=0  \implies \lim_{x \to 2} \frac{1}{x-2}=\lim_{x \to 2} \frac 4{x^2-4}$$.
But $\frac1{x-2} \neq \frac 4{x^2-4}$ or $\frac 1{x-2} \neq \frac1{x-2} \frac4{x+2}$
Thus $\lim_{x \to 2} \frac{1}{x-2}-\frac 4{x^2-4}=0$ do not true.
</s>