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

The limit ${\lim}_{x \to 0} \sin \left(\frac{1}{x}\right)$ does not exist. You can proove this taking two sequences such as ${a}_{n} = \frac{1}{2 \pi n}$ and ${b}_{n} = \frac{1}{\left(2 n + 1\right) \cdot \pi}$ and replacing into the value of x.Hence
$\sin \left(\frac{1}{\frac{1}{\left(2 \pi n\right)}}\right) = \sin \left(2 \cdot \pi \cdot n\right) = 1 \mathmr{and} - 1$.Hence the values of the limits both sequences are different such limit does not exist