# Question #e8be5

Dec 23, 2015

The limit is $0$

#### Explanation:

The easiest method to find this is to use the squeeze theorem. This theorem states the following in this case: suppose you find two functions $g \left(x\right)$ and $h \left(x\right)$ for which holds that $g \left(x\right) \le x \sin \left(\frac{1}{x}\right) \le h \left(x\right)$.
If ${\lim}_{x \to 0} g \left(x\right) = L = {\lim}_{x \to 0} h \left(x\right)$, then it definitely holds that:
${\lim}_{x \to 0} x \sin \left(\frac{1}{x}\right) = L$.

The hardest part is to find two functions for which this holds.
You can do this by noticing that the sine function can only go from $- 1$ to $1$. This means that the function is always bigger than $- 1 \cdot | x |$ and always smaller than $1 \cdot | x |$:
$- | x | \le x \sin \left(\frac{1}{x}\right) \le | x |$

Let's look at the limits of the two functions:
${\lim}_{x \to 0} - | x | = {\lim}_{x \to 0} | x | = 0$

This means that ${\lim}_{x \to 0} x \sin \left(\frac{1}{x}\right) = 0$