Answer
$$\lim_{ x \to -5} \frac{x^3+125}{x+5}=75$$
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline
x & -5.1 & -5.01 & -5.001 & -5 & -4.999 & -4.99 & -4.9 \\ \hline
f(x) & 76.51 & 75.1501 & 75.015001 & \text{undefined} & 74.985001 & 74.8501 & 73.51 \\ \hline
\end{array}$$
Work Step by Step
Looking at the graph, we can find that when $x$ approaches $-5$ from left and right, the function approaches $75$. So we can conclude that $\lim_{x \to -5} \frac{x^3+125}{x+5}$ is approximately $75$. The table also confirms our conclusion.
Now, we want to find the limit analytically. Since we get the indeterminate form $\frac{0}{0}$ by direct substitution, we should rewrite the ratio, by factoring and dividing out like factors, as$$\frac{x^3+125}{x+5}=\frac{(x+5)(x^2-5x+25)}{x+5}=x^2-5x+25$$(Please note that $x \neq -5$). Hence,$$\lim_{ x \to -5} \frac{x^3+125}{x+5}=\lim_{x \to -5} x^2-5x+25=75.$$
