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I need to prove that $f$ continuous at $(x)=0$ using a $\epsilon$-$\delta$ proof $$ f(x, y) = \begin{cases} x^2sin(\frac1x),&x\neq 0 \\ 0,&x = 0 \end{cases} $$

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it also possesses a first derivative everywhere. The derivative, however, is discontinuous at $0.$ Go figure. – Will Jagy Feb 20 '13 at 23:26
up vote 6 down vote accepted

If $0<|x|<\delta$ then $$|f(x)-f(0)| = |x^2 \sin(1/x)|\le|x|^2 < \delta^2$$ Thus we for every $\epsilon>0$ we can choose $\delta = \sqrt{\epsilon}$ such that...

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Be sure you check out the edits I made, and note why each of them--save changing $x^2$ to $|x|^2$--was essential for the proof. – Cameron Buie Feb 22 '13 at 16:06

Hint: For $x\neq 0$, $$|f(x)-f(0)|=\left|x^2\cdot\sin\frac1x-0\right|=|x|^2\left|\sin\frac1x\right|\leq|x|^2=|x-0|^2.$$

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This post is only meant to be a supplement to the answers posted above.(So that the process is understood.)

You have to prove that the limit of the function as x tends to zero is the same as the value of the function at zero.

Here epsilon is the upper bound on the distance between an abitrary f(x) and and f(0). And delta is the upper bound on the distance between an arbitrary x and 0. If you can prove that you can bring f(x) as close to f(0) as you wish, by bringing x as near to zero, you prove that the limit of the function as x tends to zero exists.(This can be done without the epsilon - delta definition, by using algebraic manipulation, but it won't be rigorous.More rigorously, for any arbitrary epsilon, you should be able to find a delta - and for this you have to use the epsilon-delta definition.)

Note that epsilon and delta are both greater than zero, so this excludes the case f(x) = f(0) and the case x = 0. So it only proves that the limit exists, it says nothing about the behaviour of the function at x= 0. So you have to prove that the limit above is equal to the value of f(x) at x = 0. Here since it is already given that f(0) = 0, all you have to do is to state the fact.(A practically trivial, but logically important step).

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We don't really need to exclude equality in either case. Suppose we're trying to do an $\epsilon$-$\delta$ proof of the continuity of a function $g:\Bbb R\to\Bbb R$ at $x=a$. Think of $\epsilon$ as the length of a chain attached to $g(x)$ and $g(a)$; $\delta$ to $x$ and $a$, where we want to make the $\delta$-chain short enough that $g(x)$ doesn't hurt itself when $x$ runs around. We will still let $x$ and $g(x)$ run around a bit, and they can even come right up to the base of the chain, i.e.: $g(x)=g(a)$ or $x=a$. – Cameron Buie Feb 21 '13 at 15:41
Of course, it isn't very interesting to let $x=a$, because then we'll necessarily have $g(x)=g(a)$. Hence, some texts require $x\neq a$. In some cases, though, that restriction won't be enough to avoid situations with $g(x)=g(a)$. For example, consider $f(x)$ from this post, and say we take $\epsilon=1$. Then taking $\delta=1$, we know for any $0<|x|<\delta$ that $|f(x)-f(0)|<\epsilon$. However, there are infinitely-many such $x$ with $|f(x)-f(0)|=0$, namely $x=\frac{1}{k\pi}$ for non-zero integers $k$, and these $x$ get arbitrarily close to (but not equal to) $0$. – Cameron Buie Feb 21 '13 at 15:43
So, even if we don't want to allow $x=a$ in our $\epsilon$-$\delta$ definition, we have to allow the possibility that $g(x)=g(a)$. – Cameron Buie Feb 21 '13 at 15:46
I am now thinking, about your sentence 'that restriction won't be enough to avoid situations with g(x) = g(a)' - a similar situation occurs in the rigorous proof of the chain rule in differentiation, when for a non- zero change in the independent variable , the change in the inner function could be zero. I was looking for an example, here I found it. Thanks once again! But do reply on my earlier post. – Nikhil Panikkar Feb 22 '13 at 12:42
If all we want to show is that $\lim_{x\to a}g(x)=L$ for some real $L$, then we certainly must require that $x\neq a$, for as you say, $g$ needn't be defined at $a$, or (even if it is defined) we may have $g(a)\neq L$ in the case that $g$ has a removable discontinuity at $a$. You're quite right about that. However, if we're trying to prove that the function $g$ is continuous at $a$, then neither of these will be concerns. (P.S.: Always glad to accidentally answer another question.) – Cameron Buie Feb 22 '13 at 14:50

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