Is this function continuous at $x = 0$?

Question

Is $$ f(x)= \begin{cases} \frac{\sin x}{|x|}, & {if}\, x\ne 0\\ 1 &, {if}\, x = 0\\ \end{cases} $$ continuous at x = 0? (This exercise is from "Calculus", 8th edition, volume 1, by Howard Anton, Irl Bivens, Stephen Davis, on page 160, exercise 48.)

Firstly, the function is defined at x = 0. Next step is to verify whether the following limit exists. $$ \lim_{x\to 0} \frac{\sin x}{|x|} $$ This is where I got stuck. I have plotted its graph on Geogebra and it was obvious that the limit has unequal one-sided limits. However, I have no idea how to make such conclusion algebraically. Thanks in advance.

Answer 1

Firstly, I think your function should be

$$f \left( x \right) = \begin{cases} \dfrac{\sin x}{\left| x \right|} & x \neq 0 \\ 1 & x = 0 \end{cases}$$

When $\lim\limits_{x \rightarrow 0^+}$, we have $\left| x \right| = x$ and for $\lim\limits_{x \rightarrow 0^-}$, we have $\left| x \right| = - x$. Thus, we get

$$\lim\limits_{x \rightarrow 0^+} \dfrac{\sin x}{\left| x \right|} = \lim\limits_{x \rightarrow 0^+} \dfrac{\sin x}{x} = 1$$

and

$$\lim\limits_{x \rightarrow 0^-} \dfrac{\sin x}{\left| x \right|} = \lim\limits_{x \rightarrow 0^-} \dfrac{\sin x}{- x} = -1$$

Therefore, the function cannot be continuous

Answer 2

As you mentioned, the right limit and left limit of$$\frac{\sin x}{|x|}$$ are not the same as x- approaches to $0$ thus $$\lim_{x\to 0} \frac{\sin x}{|x|}$$ does not exist.

Since the $$\lim_{x\to 0} \frac{\sin x}{|x|}$$ does not exist at $x=0$, the function is not continuous at $x=0.$