# Thread: Prove continuity of a fuction

1. ## Prove continuity of a fuction

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

Thanks in advance for any replies.
Feyomi.

2. Originally Posted by feyomi
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

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?

3. 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..

4. Originally Posted by feyomi
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?

5. Originally Posted by Jhevon
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?
|xsin(1/x)| = |f(x)-0| ≤ |x| = |x-0|
I would need to show that |x|<δ implies |f(x)|<ε ?
I'm sorry, Sir, I'm quite clueless as to what to do

6. Originally Posted by feyomi
|xsin(1/x)| = |f(x)-0| ≤ |x| = |x-0|
I would need to show that |x|<δ implies |f(x)|<ε ?
I'm sorry, Sir, I'm quite clueless as to what to do
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?

7. δ = ε/sin(1/x) ?

8. Originally Posted by feyomi
δ = ε/sin(1/x) ?
hehe, no. you can't have $\displaystyle x$ in it. it must be some constant dependent on $\displaystyle \epsilon$

look carefully at what i have so far, it's right there. i gave you what we can bound |f(x) - 0| with (and sin(1/x) doesn't come into it)

9. Yeah I just don't know..

10. Originally Posted by feyomi
Yeah I just don't know..
look at what we want.

we want $\displaystyle |f(x) - 0| = \bigg| x \sin \frac 1x \bigg| \le |x| < \epsilon$

and we also want $\displaystyle |x| < \delta$

you can't see a convenient $\displaystyle \delta$ to choose?

11. Oh okay so we take δ=ε ?
And then is that the end of the proof? Seeing as we've found a value of δ such that |x-0|<δ implies |f(x)-0|<ε ?

12. Originally Posted by feyomi
Oh okay so we take δ=ε ?
And then is that the end of the proof? Seeing as we've found a value of δ such that |x-0|<δ implies |f(x)-0|<ε ?
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.

13. We got there in the end haha.
Your help's been much appreciated. Thank you.