# 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 $\infty$?

• Okay...basically I realize that was not a productive way of solving this. Basically the top grows much faster than the bottom so it goes to infinity. SO I guess a lot of limits is just conceptualizing the limit...But that raises the question: when do I use 1/x rule? Jun 24 '14 at 0:23

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.
As $x\to -\infty$, we have $\frac{1}{x^3}\to 0^-$ and $\frac{1}{x}+\frac{7}{x^2}\to 0^-$ (that is, both of these expressions go to zero, but are negative, as $x$ gets larger negatively). Thus both numerator and denominator are negative, but the absolute ratio gets arbitrarily large. Thus the limit is $\infty$ rather than $-\infty$.