calculus - Find limit of $lim_{xto 0}{left(sqrt{x^6+5x^4+7x^2}cos(1+x^{-1000})right)}$, if it exists

Continuing my practice at solving limits, I'm currently trying to solve the following limit:

$$\lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)}$$

What I've done:

I know that since the cosine is an oscillating function, $\cos(\infty)$ doesn't converge to a single value, but I thought that, perhaps, by using trigonometric formulas and various tricks, I could bring the limit to a form that can be solved, but in vain.

Here is an attempt using the trigonometric formula $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$:

$$\begin{align} l & = \lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cdot \cos(1+x^{-1000})\right)}\\ & = \lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\right)}\cdot \lim_{x\to 0}{\left(\cos(1)\cos(x^{-1000})-\sin(1)\sin(x^{-1000})\right)}\\ & = 0\ \cdot \lim_{x\to 0}{\left({{\cos(1)}\over{1/\cos(x^{-1000})}}-{{\sin(1)}\over{1/\sin(x^{-1000})}}\right)}\\ \end{align}$$

Question:

Can the above limit be solved or it's simply undefined, since $\cos(\infty)$ diverges?

Answer

You are correct, the limit is $0$. Another elegant way to prove is using squeeze theorem.

Using $$-1 \le \cos (x) \le 1$$

We've

$$-\sqrt{x^6+5x^4+7x^2} \le {\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)} \le \sqrt{x^6+5x^4+7x^2}$$

Thus,

$$\lim_{x\to 0} \sqrt{x^6+5x^4+7x^2} \le \lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)} \le \lim_{x\to 0}\sqrt{x^6+5x^4+7x^2}$$

$$\implies 0\le \lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)} \le 0$$

$$\implies \color{blue}{\lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)}=0}$$