# lim_(x->0)sin(1/x)/(sin(1/x)) ?

## Find the limit ${\lim}_{x \to 0} \sin \frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}$ How would you approach this? Is it $1$ or it doesn't exist?

May 24, 2018

${\lim}_{x \rightarrow 0} \setminus \sin \frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)} = 1$

#### Explanation:

we seek:

$L = {\lim}_{x \rightarrow 0} \setminus \sin \frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}$

When we evaluate a limit we look at the behaviour of the function "near" the point, not necessarily the behaviour of the function "at" the point in question, thus as $x \rightarrow 0$, at no point do we need to consider what happens at $x = 0$, Thus we get the trivial result:

$L = {\lim}_{x \rightarrow 0} \setminus \sin \frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}$

$\setminus \setminus = {\lim}_{x \rightarrow 0} \setminus 1$

$\setminus \setminus = 1$

For clarity a graph of the function to visualise the behaviour around $x = 0$
graph{sin(1/x)/sin(1/x) [-10, 10, -5, 5]}

It should be made clear that the function $y = \sin \frac{\frac{1}{x}}{\sin} \left(\frac{1}{x}\right)$ is undefined at $x = 0$

May 24, 2018

#### Explanation:

The definitions of limit of a function I use are equivalent to:

${\lim}_{x \rightarrow a} f \left(x\right) = L$ if and only of For every positive $\epsilon$, there is a positive $\delta$ such that for every $x$, if $0 < \left\mid x - a \right\mid < \delta$ then $\left\mid f \left(x\right) - L \right\mid < \epsilon$

Because of the meaning of "$\left\mid f \left(x\right) - L \right\mid < \epsilon$", this requires that for all $x$ with $0 < \left\mid x - a \right\mid < \delta$, $f \left(x\right)$ is defined.

That is, for the required $\delta$, all of $\left(a - \delta , a + \delta\right)$ except possibly $a$, lies in the domain of $f$.

All of this the gets us:

${\lim}_{x \rightarrow a} f \left(x\right)$ exists only if $f$ is defined in some open interval containing $a$, except perhaps at $a$.

($f$ must be defined in some deleted open neighborhood of $a$)

Therefore, ${\lim}_{x \rightarrow 0} \sin \frac{\frac{1}{x}}{\sin} \left(\frac{1}{x}\right)$ does not exist.

A nearly trivial example

$f \left(x\right) = 1$ for $x$ an irrational real (undefined for rationals)

${\lim}_{x \rightarrow 0} f \left(x\right)$ does not exist.