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:
$$\begin{align}
\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})&amp;=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}\\
&amp;=\lim_{x\to2}\frac{1}{x-2}-4\lim_{x\to2}\frac{1}{x+2}\lim_{x\to2}\frac{1}{x-2}\\
&amp;=\lim_{x\to2}\frac{1}{x-2}-4\frac14\lim_{x\to2}\frac{1}{x-2}\\
&amp;=\lim_{x\to2}\frac{1}{x-2}-\lim_{x\to2}\frac{1}{x-2}\\
&amp;=0
\end{align}$$

I actually couldn't explain her why is she wrong. Cause technically it looks fine. What am i missing?
The error was in the very first step. We can say that $$\lim_{x\to a}\bigl(f(x)+g(x)\bigr)=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$$ provided that both of the limits on the right-hand side exist. In this case--$f(x)=\frac1{x-2}$ and $g(x)=-\frac4{x^2-4}$ with $a=2$--neither of these limits exist.
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.