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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.

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  • $\begingroup$ Try taking the limit from the left and the right. If we consider taking the limit from the right, then the numerator and the denominator both go to 0. Any ideas how you would deal with that? $\endgroup$ – NicNic8 Jul 11 '18 at 2:46
  • $\begingroup$ Hint: $\frac{\sin x}{|x|} = \frac{\sin x}{x} \cdot \frac{x}{|x|}$ for $x \neq 0$. $\endgroup$ – JavaMan Jul 11 '18 at 3:03
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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

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  • $\begingroup$ I completely forgot about that trigonometric limit. $\endgroup$ – Mauricio Mendes Jul 11 '18 at 2:55
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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 $x=0.$

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