Squeeze Law and Signs

Question

I am trying to find $$ \lim \limits_{x \to 0} x\cos \left(\frac2x\right) $$ through the squeeze law. To do this, I have set up the inequality $$ -1\le\cos \left(\frac2x\right)\le1 $$ Intuitively, I wrote down the next step, which would be $$ -x\le x \cos \left(\frac2x\right)\le x $$ and then taking the limits of all three. But I then realized that this inequality would not hold if $x$ was a negative number. How do I overcome this?

Answer 1

Note that

$$-\left|x\cos \left(\frac2x\right)\right|\le x\cos \left(\frac2x\right)\le \left|x\cos \left(\frac2x\right)\right|$$

and

$$0\le \left|x\cos \left(\frac2x\right)\right|\le |x|\to 0$$

then by squeeze theorem

$$\left|x\cos \left(\frac2x\right)\right|\to 0 \implies x\cos \left(\frac2x\right)\to 0$$

Answer 2

Use absolute values to avoid having to consider cases.

$$0\leq |x\cos(2/x)-0|= |x||\cos(2/x)|\leq |x|$$