Prove that f(x)={xsin(1/x), if x≠0
{ 0 , if x=0 is continuous using
i) language of limits
ii) ε - δ - language
Please help me
Thanks in advance for any replies.
Feyomi.
clearly each piece of the function is continuous when $\displaystyle x \ne 0$. to show that $\displaystyle f(x)$ is continuous then, we need only prove that $\displaystyle \lim_{x \to 0}f(x) = f(0) = 0$ (this si by the definition of continuity at a point).
for (i) use the squeeze theorem to show this. that part shouldn't be that hard.
for (ii) apply the $\displaystyle \epsilon - \delta$ definition of the limit. do you know what it is?
ah yes, i understand the question much more now.
However, is it enough the show that it is continuous at x=0?
Would I not need to prove that it is continuous otherwise?
I have just applied the Sandwich Rule (as I like to call it), but I do not know how to do the second part.
I do know the definition, but I don't know how to apply it..
to prove the function is continous, we need to prove it is continuous at all points in its domain. now, for $\displaystyle x \ne 0$, the function is a product of two functions that are continuous when $\displaystyle x \ne 0$ and hence the function is continuous when $\displaystyle x \ne 0$. the only problem is when $\displaystyle x = 0$, and so that's what we have to deal with.
as for part (ii), i will give you a hint. $\displaystyle \bigg| x \sin \frac 1x - 0 \bigg| = \bigg| x \sin \frac 1x \bigg| \le |x|$ for $\displaystyle x \ne 0$
what can you do with that?
don't worry about being lost. it was difficult for all of us the first few times.
we want to choose a $\displaystyle \delta > 0$ so that $\displaystyle |f(x) - 0|< \epsilon$ (for any $\displaystyle \epsilon > 0$) whenever $\displaystyle |x| < \delta$
so what would you choose $\displaystyle \delta$ to be?
Yes! Bingo!
Let $\displaystyle \epsilon > 0$ be given and choose $\displaystyle \delta = \epsilon$. Then, $\displaystyle 0 < |x - 0| = |x| < \delta$ implies
$\displaystyle \bigg| x \sin \frac 1x - 0 \bigg| = \bigg| x \sin \frac 1x \bigg| \le |x| < \epsilon$
Thus, $\displaystyle \lim_{x \to 0} f(x) = 0 = f(0)$ as desired.