Prove using the epsilon-delta ($\epsilon-\delta$) definition of a limit that $$\lim_{x\to2}\frac{x^2-4}{-x^2+7x-10}=\frac{4}{3}.$$
I am perplexed about how to solve this question, tried all the ways, and wasted hours and hours. Any solving is highly appreciated
First, notice that, for $x\notin\{2,5\}$, we have that
$$\frac{x^2-4}{-x^2+7x-10}=\frac{(x-2)(x+2)}{-(x-2)(x-5)}=-\frac{x+2}{x-5}.$$
Furthermore
$$\left\lvert-\frac{x+2}{x-5}-\frac{4}{3}\right\rvert=\frac{7}{3}\left\lvert\frac{x-2}{x-5}\right\rvert.$$
Now if we restrict ourselves to that we have that $\lvert x-2\rvert<1$, then it is easy to check that we also have that $\lvert x-5\rvert>1$. Set
$$\delta=\min\left\{1,\frac{3}{7}\varepsilon\right\}.$$
Then, if $\lvert x-2\rvert<\delta$, it follows that
$$\frac{7}{3}\left\lvert\frac{x-2}{x-5}\right\rvert<\varepsilon.$$
Combining this with the first equalities, the result follows.