Squeeze Theorem: Finding the limit of a trig function I'm stuck on finding the limit of a complex fraction/trig function.  Could someone please assist, or point out where I'm going wrong?

Determine
$$\lim\limits_{x \to 0} \frac{(x+1)\cos(\ln(x^2))}{\sqrt{(x^2+2)}}$$

For all $x$: $$-1 \le \cos(\ln(x^2)) \le 1$$
Multiply by $(x+1)$:
$$-(x+1) \le (x+1)\cos(\ln(x^2)) \le (x+1)$$
Divide by $\sqrt{(x^2+2)}$:
$$\frac{-(x+1)}{\sqrt{(x^2+2)}} \le \frac{(x+1)\cos(\ln(x^2))}{\sqrt{(x^2+2)}} \le \frac{(x+1)}{\sqrt{(x^2+2)}}$$
Now to find the limits:
$$\lim\limits_{x \to 0} \frac{-(x+1)}{\sqrt{(x^2+2)}} = \frac{-1}{\sqrt{(2)}}$$
$$\lim\limits_{x \to 0} \frac{(x+1)}{\sqrt{(x^2+2)}} = \frac{1}{\sqrt{(2)}}$$
This is where I get stuck.  My limits are not equal, so I cannot solve using the Squeeze Theorem.  I must've skipped a step or used the wrong approach, but I'm not sure where.

$$f(x)=\frac{(x+1)\cos(\ln(x^2))}{\sqrt{x^2+2}}$$
Take two sequences tending to $0$:
$$\begin{array}{}
a_n=\exp\left(\tfrac{\pi}{4}(1-2n)\right),&\lim\limits_{n\to\infty} a_n=0\\
b_n=\exp(-n\pi),&\lim\limits_{n\to\infty} b_n=0
\end{array}$$
We have
$$f(a_n)=\frac{(a_n+1)\cos\left(\ln\left(\exp\left(\tfrac{\pi}{4}(1-2n)\right)^2\right)\right)}{\sqrt{a_n^2+2}}
=\frac{(a_n+1)\cos\left(\tfrac{\pi}{2}(1-2n)\right)}{\sqrt{a_n^2+2}}=0$$
$$f(b_n)=\frac{(x+1)\cos(\ln(\exp(-n\pi)^2))}{\sqrt{x^2+2}}
=\frac{(x+1)\cos(-2n\pi)}{\sqrt{x^2+2}}
=\frac{b_n+1}{\sqrt{b_n^2+2}}$$
Calculating the limits of $f(a_n)$ and $f(b_n)$ we get:
$$\lim_{n\to\infty}f(a_n)=0$$
$$\lim_{n\to\infty}f(b_n)=\tfrac{\sqrt2}{2}$$
Thus the limit
$$\lim_{x\to0}f(x)$$
doesn't exist.
