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.

  • 3
    $|f(x)-0|=|f(x)|$, but this isn't true if you replace $0$ by $l$. – vadim123 Aug 2 '13 at 23:29
  • 2
    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. – user84413 Aug 2 '13 at 23:33
  • @user84413, why? That is only the case if he wants to show that $f(x)$ is uniformly continuous, if I'm not mistaken. – Alex Wertheim Aug 2 '13 at 23:37
  • Yes, I don't want to show that f(x) is uniformly continuous. – mauna Aug 2 '13 at 23:53
  • 1
    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 $x$, $|x - a| < \delta \implies |f(x) - L| < \epsilon$. – Pratyush Sarkar Aug 3 '13 at 2:49
up vote 1 down vote accepted
+50

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:

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

so:

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

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 $G(0)\not=0$.

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.

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

$|L-M|=|(L-f(x))+(f(x)-M)|\le|L-f(x)|+|f(x)-M|<\epsilon+\epsilon=2\epsilon=\frac{|L-M|}{50}$

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 $0$.

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.
  • 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. – zyx Aug 10 '13 at 18:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.