Evaluating a limit as $x \to -\infty$ of a power of a rational function Sorry for the weird title, I don't know how to put the equation on the title.

$$\lim_{x\to-\infty}\left(\frac{1-x^3}{x^2+7x}\right)^5$$

Ok I divided inside the parenthesis by $x^2$, but now I am stuck.
If we divide numerator and denominator in the parentheses by $x^3$ we get

$$\frac{1-x^3}{x^2+7x}=\frac{ 1/x^3 - 1}{1/x + 7/x^2}$$

Now note that as $x\to -\infty$ the numerator approaches $-1$ and the denominator approaches $0$. So overall the fraction diverges.
This already shows that the original limit diverges since the exponential function is continuous.
Since $X^5$ is continuous, it suffice to calculate $$\lim_{x\to-\infty}\left(\frac{1-x^3}{x^2+7x}\right)$$. Then by using L'Hospital's rule twice, you get the answer $\infty$.

Without using LHospital's rule, divide $x^2$ on the numerator and denominator.
$$\lim_{x\to-\infty}\left(\frac{\frac{1}{x^2}-x}{1+\frac{7}{x}}\right)$$

As $x\to -\infty$, $\frac{1}{x^2}\to 0, \frac{1}{x}\to 0$, hence numerator $\to \infty$, denominator $\to 1$, so the limit is $\infty$