# How do you find the limit of sinx / x as x approaches oo?

Mar 29, 2016

We use squeeze theorem, which says that for three functions $g \left(x\right) , f \left(x\right) , \mathmr{and} h \left(x\right)$,
If $g \left(x\right) \le f \left(x\right) \le h \left(x\right) , \mathmr{and} {\lim}_{x \to a} g \left(x\right) = {\lim}_{x \to a} h \left(x\right) = L$
then ${\lim}_{x \to a} f \left(x\right) = L$.

${\lim}_{x \to \pm \infty} \sin \frac{x}{x} = 0$

#### Explanation:

Given that $f \left(x\right) = \sin \frac{x}{x}$
We know that $\sin x$ oscillates between $- 1 \mathmr{and} + 1$ for all values of $x$. Therefore, if we set $g \left(x\right) = \frac{- 1}{x} \mathmr{and} h \left(x\right) = \frac{+ 1}{x}$ we have located the two functions satisfying the first condition that
$g \left(x\right) \le f \left(x\right) \le h \left(x\right)$ which can be written as

Given that (-1)/x<= sinx/x <= (+1)/x "for all values of " x " in " (-oo,oo)
We know that ${\lim}_{x \to \pm \infty} \frac{- 1}{x} = 0 \mathmr{and} \text{also that} {\lim}_{x \to \pm \infty} \frac{+ 1}{x} = 0$.

It follows from the squeeze theorem that
${\lim}_{x \to \pm \infty} \sin \frac{x}{x} = 0$