Let: $$ G(x) = \left\{ \begin{array} {cc} x \sin \frac{1}{x} , & x\neq 0 \\ 0, & x=0 \end{array} \right. $$

I can understand that the function is continuous at $x=0$ because:

For $\epsilon>0$ and $\delta>0$, this implies, for all $x$

$$ |x-0|<\delta \implies |f(x)-0| < \epsilon \\ |x|<\delta \implies |f(x)| < \epsilon \\ |x|<\delta \implies |x\sin \frac{1}{x}| < \epsilon \\ |x|<\delta \implies |x||\sin \frac{1}{x}| < \epsilon \\ \therefore \delta = \frac{\epsilon}{\sin \frac{1}{x}} $$

So, for any $\epsilon>0$, we can find a $\delta=\frac{\epsilon}{\sin \frac{1}{x}}$ so that $ |x-0|<\delta \implies |f(x)-0| < \epsilon$ is true.

Hence, $\lim \limits_{x\to 0}G(x)=G(0)=0$

My text also mentions that $G$ will only be continuos at $0$ if $G(0)=0$.

So now I am wondering why can't it be any number, $l$?

So far here is what I have come up with using the same manipulations as above:

$$|x-0|<\delta \implies |f(x)-l| < \epsilon \\ |x|< \frac{\epsilon + l}{\sin \frac{1}{x}} $$

I had expected this to lead to a contradiction when $l \neq 0$ but so far I can't see it. How do I show that $\lim \limits_{x \to 0}G(x)$ must be $0$ and where did I go wrong in my workings?

Thank you in advance for any help provided.

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    $\begingroup$ $|f(x)-0|=|f(x)|$, but this isn't true if you replace $0$ by $l$. $\endgroup$ – vadim123 Aug 2 '13 at 23:29
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    $\begingroup$ There are some things you want to change: You want to start with an $\epsilon$, and then find out what the corresponding value of $\delta$ would be; and your $\delta$ should not involve x at all. $\endgroup$ – user84413 Aug 2 '13 at 23:33
  • $\begingroup$ @user84413, why? That is only the case if he wants to show that $f(x)$ is uniformly continuous, if I'm not mistaken. $\endgroup$ – Alex Wertheim Aug 2 '13 at 23:37
  • $\begingroup$ Yes, I don't want to show that f(x) is uniformly continuous. $\endgroup$ – mauna Aug 2 '13 at 23:53
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    $\begingroup$ The reason $\delta$ can't involve $x$ is because in the rigorous definition of limits, it wouldn't logically make sense. The $\delta$ is chosen before the $x$. The definition is: $L$ is the limit of $f$ at $a$ if, for all $\epsilon > 0$, there is a $\delta > 0$, so that for all $x$, $|x - a| < \delta \implies |f(x) - L| < \epsilon$. $\endgroup$ – Pratyush Sarkar Aug 3 '13 at 2:49

First of all, there is a caveat about your proof, the value you choose for $\delta$ can get very big or not even be a number, for example lets fix $x$ such that $\sin{\frac{1}{x}}=0$ (which actually happens an infinite number of times as x approaches to 0), then your $$ \delta=\frac{\varepsilon}{\sin{\frac{1}{x}}}, $$ is not well defined.

However, your proof is not all wrong you are just picking the wrong delta, in this case, as your function $G$ is multiplied by the well behaved identity function you can have $\delta=\varepsilon$, so the proof that $G$ is continuous at $x=0$ will be as follows:

Let $\varepsilon>0$ given and let $\delta=\varepsilon$, then for any $x$ such that: $$ |x-0|<\delta $$ we have that:

$$ |G(x)-G(0)|=|x\sin{\frac{1}{x}}-0|\\ =|x\sin{\frac{1}{x}}|\\ =|x||\sin{\frac{1}{x}}|\\ \le|x|\\ \lt\delta=\varepsilon $$ where your missing step follows from the fact that $|sin(y)|\le1$ for all $y\in\mathbb{R}$, and the triangle inequality. So we conclude that $G$ is continuous at $x=0$.

Now if you want to see why any other choice of $l$ will fail, your assumption of using a proof by contradiction is correct, nevertheless you are doing your algebra wrong.

Lets suppose there is an $l\not=0$ such that $G(0)=l$ is continuous at $x=0$. This means that for every $\varepsilon>0$ we can find a $\delta_\varepsilon>0$ (the subscript is just o make emphasis on the fact that the choice of this $\delta$ depends on the $\varepsilon$ given) such that:

$$ |x-0|<\delta_\varepsilon \Rightarrow |G(x)-G(0)|<\varepsilon. $$

So the proof goes like this:

Let $\varepsilon>0$ given, and $\delta_\varepsilon$ such that:

$$ |x-0|<\delta_\varepsilon \Rightarrow |G(x)-G(0)|<\varepsilon. $$

To make our life easier, lets take another $\delta$ to be: $\delta=\min\{\delta_\varepsilon,\varepsilon\}$, (Note that $|x-0|<\delta<\delta_\varepsilon$ so the conclusion of our assumption still holds.) hence:

$$ |x-0|<\delta, $$ then:

$$ \left||G(x)|-|G(0)|\right|\le|G(x)-G(0)|<\varepsilon, $$


$$ \left||x\sin{\frac{1}{x}}|-|l|\right|<\varepsilon, $$

it follows that:

$$ -\varepsilon<|x\sin{\frac{1}{x}}|-|l|<\varepsilon, $$

in particular, using the left side:

$$ |l|-\varepsilon<|x\sin{\frac{1}{x}}| $$

but remember that $|x\sin{\frac{1}{x}}|\le|x|$ so:

$$ |l|-\varepsilon<|x\sin{\frac{1}{x}}|\le|x|=\delta\le\varepsilon. $$

so we conclude that:

$$ |l|<2\varepsilon. $$

Which is clearly a contradiction because $l$ is fixed, and this should hold for any value of $\varepsilon$.

So $G$ is not continuous at $x=0$ if $G(0)\not=0$.

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Hint Assume that for some $l$ the function $$H(x) = \left\{ \begin{array} {cc} x \sin \frac{1}{x} , & x\neq 0 \\ l, & x=0 \end{array} \right.$$

Then, the function

$$H(x)-G(x) = \left\{ \begin{array} {cc}0 , & x\neq 0 \\ l, & x=0 \end{array} \right.$$

is continuous at $0$ as the difference of two continuous functions.

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Your text is correct. The function is continuous only if we define $G(0)=0$. This is because to get continuity of $G$ at $0$ we need that $G(0)=\lim_{x\to0}G(x)$ and limits of functions at some point in $\mathbb{R}$ (provided the limit exists) are unique when we determine convergence by means of the $\epsilon-\delta$ method.

To see this suppose that $f$ is a function defined on a neighbourhood of $a$ such that $\lim_{x\to a}f(x)$ exists where $a\in\mathbb{R}$. I claim the limit is unique.

Suppose not. Then there are two distinct numbers $L,M\in\mathbb{R}$ such that $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}f(x)=M$. This means that if we take $0<\epsilon=\frac{|L-M|}{100}$ there exists $0<\delta_{1}$ and $0<\delta_{2}$ such that if $0<|x-a|<\delta_{1}$ then $|f(x)-L|<\epsilon$ and if $0<|x-a|<\delta_{2}$ then $|f(x)-M|<\epsilon$. If we take $\delta=\min\{\delta_{1},\delta_{2}\}$ then if $0<|x-a|<\delta$ then


by the definition we gave fo $\epsilon$. But this implies $49|L-M|<0$. But this is impossible. Contradiction.

Hence limits must be unique. So the reason that we can't define $G(0)$ as anything else is because if the limit exists then it is unique. Since $|x\sin\big(\frac{1}{x}\big)|\le|x|$ then $lim_{x\to0}G(x)$ exists and is $0$.

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The direct question was not answered:

I had expected this to lead to a contradiction when $l \neq 0$ but so far I can't see it. How do I show that $\lim \limits_{x \to 0}G(x)$ must be $0$ and where did I go wrong in my workings?

You cannot get a contradiction because your workings are (with some minor re-arrangement) essentially correct.

You did not get a proof that the function with $G(0)=L$ is continuous, because for $L \neq 0$ those workings are proving something slightly different than continuity.

The endpoint of your reasoning with inequalities would be a correct proof of the true statement that when $G(0)$ is assigned the value $L$, then: for all $\epsilon > 0$, there exists a $\delta > 0$ such that:

$|G(x)| < |L| + \epsilon \quad$ when $|x| < \delta$.

For continuity with $G(0)=L$, that phrase should be replaced by

$L - \epsilon < G(x) < L + \epsilon \quad$ when $|x| < \delta$.

If $L=0$ the two phrases have the same meaning, so what your argument would demonstrate is that the function with $G(0)=0$ is continuous. For other values of $L$, it is a proof of the upper bound, but the lower bound needed to get a proof of continuity is false. There is no contradiction, because there is no proof of the false statement.

The re-arrangements of the proof needed for full correctness are:

  • replace sin by |sin| in the inequalities
  • bound |sin| by $1$, as suggested by several people
  • which leads to $\delta = \epsilon$ as the value to use in the proof
  • present the inequalities starting in the opposite order, showing that if $\delta = \epsilon$, then the conditions for continuity (with $G(0)=0$) are satisfied.
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  • $\begingroup$ Of course this is related to the uniqueness of limits, but arguments that are correct for all $L$ do not become false just because the question asked why they could not be used to prove continuity for $L \neq 0$. Rather, the arguments are correct, but only for $L=0$ do they show continuity, so there was only a misunderstanding of what the calculations were accomplishing. $\endgroup$ – zyx Aug 10 '13 at 18:38

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