1. Identify the function. 2. Find bounding functions. 3. Show bounding functions have the same limit. 4. Conclude the limit of the original function.

1. Recognize $-1 \leq \cos(\frac{1}{x}) \leq 1$. 2. Multiply by $x$: $-|x| \leq x\cos(\frac{1}{x}) \leq |x|$. 3. $\lim_{x \to 0} -|x| = 0$ and $\lim_{x \to 0} |x| = 0$. 4. Therefore, $\lim_{x \to 0} x\cos(\frac{1}{x}) = 0$.

Use absolute values to ensure inequalities hold for both positive and negative x values.

1. Find $\lim_{x \to a} g(x)$ and $\lim_{x \to a} h(x)$. 2. If both limits are equal to $L$, then $\lim_{x \to a} k(x) = L$.

If $f(x) leq g(x) leq h(x)$ and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} g(x) = L$.

The value that a function approaches as the input approaches a specific value.

Functions that enclose another function, used in the Squeeze Theorem.

A function $f(x)$ is continuous at $x=a$ if $\lim_{x \to a} f(x) = f(a)$.

If $f(x) \leq g(x) \leq h(x)$ for all $x$ in an interval containing $a$ (except possibly at $a$ itself) and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} g(x) = L$.