Proving $\lim_{x\to-\infty}x^{-k} = 0$

Question

Please prove $\lim_{x\to-\infty}x^{-k} = 0$ for k a natural number using the definition of this limit. I am having problems discovering the proof.

Answer 1

For all $k > 0$ there exists a $p\in\mathbb{N}$ such that $\frac{1}{p} < k$. Then

$$0 < \frac{1}{x^k} = \left(\frac{1}{x}\right)^k < \left(\frac{1}{x}\right)^{1/p}$$

Since $\frac{1}{x}\to 0$ and $f(u) = u^{1/p}$ is continuous at $0$, we have

$$\lim_{x\to -\infty}\left(\frac{1}{x}\right)^{1/p} = \lim_{x\to -\infty}\left(\frac{1}{x}\right)^{1/p} = \lim_{u\to 0}u^{1/p} = 0^{1/p} = 0$$

Therefore

$$\lim_{x\to -\infty}\frac{1}{x^k} = 0 \ \ k > 0$$

Answer 2

The limit is equivalent to $\lim_{x\to -\infty} \frac{x^k}{x^{2k}}$, an indeterminate form of $\frac{\infty}{\infty}$. By L'Hospitals rule, that is equivalent to $\lim_{x\to -\infty} \frac{kx^{k-1}}{2kx^{2k-1}}$. A little bit of algebraic manipulation shows that that is equal to $\lim_{x\to -\infty} \frac{1}{2}*x^{-k}$. Taking the 1/2 out of the limit, we have $\lim_{x\to -\infty} x^{-k}=\frac{1}{2}*\lim_{x\to -\infty} x^{-k}$. Treating $\lim_{x\to -\infty} x^{-k}$ as a variable in a linear equation to be solved for, we have $\lim_{x\to -\infty} x^{-k}=0$