# How do you use the Squeeze Theorem to find lim xcos(1/x) as x approaches zero?

Use the fact that the cosine function is always between $- 1$ and $1$, implying that the given function is always between $- | x |$ and $| x |$, which both go to zero as $x$ goes to zero.
Let $f \left(x\right) = x \cos \left(\frac{1}{x}\right)$, $g \left(x\right) = - | x |$, and $h \left(x\right) = | x |$. Since $- 1 \le q \cos \left(\frac{1}{x}\right) \le q 1$ for all $x \ne 0$, it follows that $g \left(x\right) \le q f \left(x\right) \le q h \left(x\right)$ for all $x \ne 0$.
But ${\lim}_{x \to 0} g \left(x\right) = {\lim}_{x \to 0} h \left(x\right) = 0$. Therefore, the Squeeze Theorem can be use to conclude that ${\lim}_{x \to 0} f \left(x\right) = {\lim}_{x \to 0} x \cos \left(\frac{1}{x}\right) = 0$.