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

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?

• By your logic, $\lim\limits_{x\to\infty}{1\over x}\cdot x$ would be 0. But there is a catch. – Ivan Neretin Aug 14 at 12:30
• The mistake is that you are first calculating a limit of a part of the whole function and then you implement that answer in the rest of the limit. So you are not implementing the limit on all components at the same time (!!) and this kind of "preferential treatment" poses a problem. What I mentioned in general, Ivan pointed that out in a specific example. – imranfat Aug 14 at 12:32
• Perhaps Limit of a product is product of limits may be helpful to look at. – Hendrix Aug 14 at 12:34

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.

• Some students also make the OP's mistake when it does not involve an indeterminate form. Even then the limit's answer can then change. In such a case it becomes harder to understand for the novice what went wrong. Usually it boils down to incorrectly applying limit laws. I have had students ignoring a cosine term in the given problem (for a limit going to 0), because, well what is a $1$ going to do? – imranfat Aug 14 at 12:40
• The full answer is $\frac{1}{(0)\infty}$ then right? $\frac{1}{(0)\infty}$ must be an indeterminate form because the value of the limit is defined (it's $-1$). But $\frac{1}{(0)\infty}$ does not match any one of the seven indeterminate forms. Why? – user532874 Aug 14 at 12:57
• @user532874 Well, it contains an indeterminate form. – Parcly Taxel Aug 14 at 12:58
• So whatever contains an indeterminate form is basically indeterminate (we can't know the value precisely but it exists) so why are there just 7 forms? – user532874 Aug 14 at 13:02
• @user532874 Those are merely seven basic forms, and any expression containing them inherits their indeterminateness. – Parcly Taxel Aug 14 at 13:03

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

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

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