Answer
$$\lim_{x \to -4}(x^2+4x)=0$$For the proof by using the $\epsilon - \delta$ definition, see below.
Work Step by Step
$$\lim_{x \to -4} (x^2+4x)=(-4)^2+4(-4)=0$$Now, we want to prove this limit by using $\epsilon - \delta$ definition; that is, we must show that for each $\epsilon >0$, there exists a $\delta >0$ such that $|(x^2+4x)-0|< \epsilon$ whenever $|x-(-4)|< \delta$.
Now, we have$$|x^2+4x|=|x||x+4|.$$For all $x$ in the interval $(-5,-3)$, $|x|<5$. So, letting $\delta$ be the minimum of $1$ and $\frac{\epsilon }{5}$, we conclude that$$|x-(-4)|< \delta =\min \{ 1, \frac{\epsilon }{5} \} \quad \Rightarrow \quad |(x^2+4x)-0|=|x||x+4|$$.
