# Verification Proof of Discontinuity of Sine Function at x=0

Show that the following function is not continuous at 00.

f(x)={0,when x=0sin(12x),when x≠0f(x) = \begin{cases}0, & \text{when $x=0$} \\ \sin\left(\frac{1}{2x}\right), & \text {when $x\neq0$} \end{cases}

Proof.

To prove discontinuity we need to analyze the One-Sided Limits of the function.

To prove that the right limit does not exist,

let consider the sequence {xn}=1(2n+1)π\{x_n\} = \frac{1}{(2n+1)\pi},

and observe that sin(12x)\sin (\frac{1}{2x}) = (−1)n(-1)^n.

Since {xnx_n} converges to 0 but (−1)n(-1)^n does not converge,

it follows from the Sequential Characterization of Continuity Theorem

that the right limit does not exist.

To prove that the left limit does not exist,

let consider the sequence {xn}=−1(2n+1)π\{x_n\} = \frac{-1}{(2n+1)\pi},

and observe that sin(12x)=(−1)n+1\sin \left(\frac{1}{2x}\right) = (-1)^{n+1}.

Since {xn}\{x_n\} converges to 00 but (−1)n+1(-1)^{n+1} does not converge,

it follows from the Sequential Characterization of Continuity Theorem that the left limit does not exist.

Therefore, f(x)f(x) is discontinuous at x=0x=0. Q.E.D.

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1

This is indeed a correct argumentation. Although actually showing that one of the one-sided limits is non zero would have been sufficient.
– b00n heT
2 days ago

Great proof! As is mentioned in the comment above, once you show the right limit doesn’t exist, you can stop there. No matter what result you get for the left limit, the left limit can’t equal the right limit since the right limit doesn’t exist, and hence the limit doesn’t exist.
– user46944
2 days ago

– Abdallah Hammam
2 days ago

1

In addition to my previous comment, an equivalent characterization of continuity of a function ff at xx is that for every sequence xnx_{n} such that xn→xx_{n} \to x, we get f(xn)→f(x)f(x_{n}) \to f(x). So you could have started the proof saying “we will exhibit a sequence which converges to x=0x=0, but for which the sequence of images doesn’t converge to f(0)=0f(0)=0”, and then your argument for the left limit would work perfectly.
– user46944
2 days ago

@b00nheT Thank you for your comment. I used both limits because there is a definition of continuity that requires only right side or left side limit, but not both.
– Beginner
2 days ago

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1

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Way overwritten. Here is a shorter proof.

There exists a sequence xnx_{n} of positive numbers that converges to 00 and such that the sequence sin(1/xn)=(−1)n\sin(1 / x_{n}) = (-1)^n fails to converge. Thus, function f(x)f(x) fails to be right-continuous at 00. Since function sin(x)\sin(x) is odd, the sequence −xn-x_{n} analogously shows failure of left-continuity at 00 for f(x)f(x).

1

I think you’re expecting too much from a beginner. OP will work their way to writing shorter proofs as they continue to learn. For their current level, their proof is fine.
– user46944
2 days ago

You are probably right. Just wanted to guard them against the Bourbakist habit of over-formalizing everything, so prevalent in modern mathematics.
– user8960
2 days ago

@user8960 Thank you so much for your ideas. I love short proofs! I just did not see a way to do it.
– Beginner
2 days ago

1

@user8960 Thank you! Very interesting!
– Beginner
2 days ago

1

Here’s an even shorter proof ;). Since f(x)=1f(x) = 1 and −1-1 in every neighbourhood of 00, ωf(0)=2≠0\omega_f(0) = 2 \neq 0, QED
– MathematicsStudent1122
2 days ago