Finding $ \lim_{x \to - \infty} \left( \frac{1 - x^{3}}{x^{2} + 7 x} \right)^{5} $. I have to evaluate the following limit:
$$
\lim_{x \to - \infty} \left( \frac{1 - x^{3}}{x^{2} + 7 x} \right)^{5}.
$$
What I did was to divide both the numerator and the denominator of the expression inside the parentheses by $ \dfrac{1}{x^{3}} $ to get
$$
\left[
\frac{\left( \dfrac{1}{x^{3}} - 1 \right)}{\left( \dfrac{1}{x} + \dfrac{7}{x^{2}} \right)}
\right]^{5}.
$$
As $ x \to - \infty $, each term in the expression above goes to $ 0 $ except for the $ -1 $.

Question: Why is the final limit $ \infty $?


Consider dividing top and bottom of $\dfrac{1-x^3}{x^2+7x}$ by $x^2$. We get 
$$\frac{\frac{1}{x^2}-x}{1+\frac{7}{x}}.$$
Now let $x$ get large negative. It is clear that the bottom approaches $1$, and that the top blows up. 
