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$$f(x) = \begin{cases} \frac{\sin x}{|x|} &\text{ if }x \neq0 \\ \hspace{0.3cm}1 &\text{ if }x=0. \end{cases}$$

My Attempt

1)$$\lim_{x\rightarrow0}\frac{\sin x}{|x|} = \lim_{x\rightarrow0}\frac{\sin x}{x} \frac{x}{|x|} = 1\lim_{x \rightarrow0}\frac{x}{|x|} \\$$ 2)$$\lim_{x\rightarrow0^{-}}\frac{x}{|x|}=-1 \hspace{0.3cm}\text{and}\hspace{0.3cm} \lim_{x\rightarrow0^{+}}\frac{x}{|x|}=1 $$ Therefore: $$1\lim_{x\rightarrow0}\frac{x}{|x|}=DNE $$ so, $f$ is not continuous at $0$.

My question is does my solution actually prove that $f$ is not continuous at $0$? or is it continuous at zero because $f(x)=1$ when $x=0$?

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    It's not continuous at $x=0$, but you can call it *semi*-continuous there. See https://en.wikipedia.org/wiki/Semi-continuity – 2017-02-16
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    To be continuous f(x) most equal the limit. If it doesn't have a limit it can't equal it. f(x) = 1 $\ne \lim_{x\rightarrow0^{-}}\frac{x}{|x|}$ so there is nothing more to be said. – 2017-02-16
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    @BarryCipra thats a good point on the interval I = [0, infinity) yes? – 2017-02-16
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    @NickPavini, I'm not sure what you want my assent to. If you draw the graph of your function $f$, you'll see it has what's called a *jump discontinuity* at $x=0$. It's "continuous from the right" but not from the left (which is what your two evaluations for $x\to0^+$ and $x\to0^-$ showed). – 2017-02-16

4 Answers 4

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1)$$\lim_{x\rightarrow0}\frac{\sin x}{|x|} = \lim_{x\rightarrow0}\frac{\sin x}{x} \frac{x}{|x|} = 1\lim_{x \rightarrow0}\frac{x}{|x|} \\$$

Note that the second equality does not hold, since for two sequences $(a_n)$, $(b_n)$ you only have $$ \lim_{n \to \infty} a_n b_n = \lim_{n \to \infty} a_n \lim_{n \to \infty} b_n $$ provided that both sequences converge. Hence in your case it would be better to start directly with a modification of part 2) :

If $f(x)$ is continuous at $0$ then the following equality is necessary: $$ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x). $$ But on the one hand you have $$ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin(x)}{|x|} = \lim_{x \to 0^+} \frac{\sin(x)}{x} = 1 $$ and on the other hand $$ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(x)}{|x|} = \lim_{x \to 0^-} \frac{\sin(x)}{-x} = -\lim_{x \to 0^+} \frac{\sin(x)}{x} = -1. $$ So $f$ can't be continuous at $0$.

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You made a confusion between constant at $0$ and continuous at $0$.

Basically, you just have to say that

$$\lim_{x\to 0^-}\frac{\sin x}{\vert x\vert}=\lim_{x\to 0^+}\frac{-\sin(x)}x=-1$$

but $f(0)=1$ so $f$ is not continuous at $0$.

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    Ok i see but the limit from the left is 1 where the limit from the right is -1.. does proving the limit does not exist suffice? Or do i need to reference $f(0)$? – 2017-02-16
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    @NickPavini It is sufficient. Both ways of proving that result are correct! – 2017-02-16
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    Two things can not be equal if one of them does not exist. No point in talking about what the thing that does exist *is* as it doesn't matter... because it can't be equal to the thing that does not exist. – 2017-02-16
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You showed that it is not continuous, not that it is not constant. Very different concepts.

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    Thank you your right my mistake – 2017-02-16
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    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - [From Review](/review/low-quality-posts/763302) – 2017-02-16
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    @zhoraster, check the edit history. – 2017-02-16
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Sin(x)/|x| So for x>0 lim(x--->0). Sin(x)/x On applying limit comes out to be 1

While for x<0 Lim(x---->0) sin(x)/(-x) By applying limit we get anwer -1 Therefore as left hand limit is not equal to right hand limit so f(x) is discontinuous as f(x) =1 at x=0

Any kind of query in my answer is welcomed also you can inform me of any mistake in my attempt if any.