what does it mean if after factoring limit have $\left(\frac{0}{0}\right)$
for example $\lim _{x\to -1}\left(\frac{x^2+2x-8}{x^2+5x+4}\right)$
If plug in $-1$ that will get $=-\frac{9}{0}$ which is undefined
and if factor that will get $\frac{\left(x+4\right)\left(x-2\right)}{\left(x+4\right)\left(x+1\right)}$ = $\frac{\left(x-2\right)}{\left(x+1\right)}$ = $\frac{\left(-1-2\right)}{\left(-1+1\right)}=-\frac{3}{0}$
what does that mean? what should I do now?
The limit for $x \to -1$ is not undefined. The two forms $\dfrac{-9}{0}$ and $\dfrac{-3}{0}$ suggest that the limit is $\infty$ and an inspection of the signs of the function shows that there are two different limits $$ \lim_{x \to -1^-} f(x)= +\infty \quad and \quad \lim_{x \to -1^+} f(x)= -\infty $$
Your decomposition is correct, and shows that the limit for $x \to -4$ is indeterminate. Here we have a form $\dfrac{0}{0}$, the function is not defined, but the discontinuity is eliminable since $$ \lim_{x \to -4} f(x)= 2 $$
