Evaluating $\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$

Question

Problem:

$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$

$$\lim_{x \to -\infty} \sqrt{\frac{1}{x^6}}=0$$ so... $$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}=\frac{1}{0}$$

The answer is $-1$ and I know how to get that answer. Where is the mistake in this method though?

Answer 1

Just as $\sqrt{1/x^6}$ goes to $0$, so does $\sqrt{x^6+4}$ go to $\infty$. You cannot substitute just one of these radicals and then simplify, and their unsimplified product is the indeterminate form $0\cdot\infty$ and so cannot be handled directly.

Answer 2

Yes, $\lim_{x\to-\infty}\sqrt{\frac{1}{x^6}}=0$. But $\lim_{x\to-\infty}\sqrt{x^6+4}=\infty$.

Answer 3

It is $$\lim_{x\to \infty}\frac{1}{-\sqrt{1+\frac{4}{x^6}}}=…$$

Answer 4

$\frac 1 {-\sqrt {\frac 1 {x^{6}}} \sqrt {x^{6}+4}}$ is nothing but $\frac 1 {- \sqrt {\frac 4 {x^{6}}+1}}$ so the limit is $\frac 1 {-1}=-1$.