Find limit of $\lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)}$, 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?
 A: 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, 
$$\lim_{x\to 0} \sqrt{x^6+5x^4+7x^2} \le \lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)} \le \lim_{x\to 0}\sqrt{x^6+5x^4+7x^2}$$
$$\implies 0\le \lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)} \le 0$$
$$\implies \color{blue}{\lim_{x\to 0}{\left(\sqrt{x^6+5x^4+7x^2}\cos(1+x^{-1000})\right)}=0}$$
A: Correct me if wrong :
For $|x| \lt 1:$
$ 0\le  |\sqrt{(x^6+5x^4+7x^2)} |\cdot $
$|\cos(1+x^{-1000})|\le 4|x|.$
The limit $x \rightarrow 0$ is?
Used : 
1) $|\cos(\theta) | \le 1.$
2) $x^6 + 5x^4+ 7x^2 \le $
$x^2 + 5x^2+ 7x^2 \le 16x^2$ for $|x| \lt 1.$
A: In $x\sqrt{x^4+5x^2+7}\cos(1+x^{-1000})$, the first factor tends to zero, the second to a finite value ($7$) and the third remains bounded (in $[-1,1]$).
Hence the limit is zero.
