Question

HELP! - prove that the function x sin 1/x is continuous for every value in the range of (0,1)

Answer 1

using what? it is the composite of two continuous functions (well \(\frac{1}{x}\) is continuous on its domain) so it must be continuous

Answer 2

using epsilon and delta.

Answer 3

I think I have to prove it using for every d\[\delta >\ \left| x2-x1 \right| \] there exists also \[\epsilon > \left| f(x2)-f(x1) \right|\]

Answer 4

Other way around. For every \(\varepsilon >0\) there exists a \(\delta >0\) so that \(|f(x)-f(x_0)|<\varepsilon\) for all \(x\) with \(|x-x_0|<\delta\).

Think about it this way: \(\delta\) is a function of \(\varepsilon\).

Answer 5

You should get someone else to help you, I'm sort of busy right now.

Like this:
@mukushla

Answer 6

Yes, I really need help on this. I have ten questions that I need to solve by tomorrow or I will probably not pass on my course :(

Answer 7

ok this is not so bad at all because the largest sine can be is 1

Answer 8

let \(\epsilon>0\) be given and you want a \(\delta\) which you write in terms of \(\epsilon\) so that if \(|x-y|<\delta\) then \(x\sin(\frac{1}{x})-y\sin(\frac{1}{y})|<\epsilon\)

Answer 9

last line is a typo
should be
\[|x\sin(\frac{1}{x})-y\sin(\frac{1}{y})|<\epsilon\]

Answer 10

ok what's next? How do I get from here to the proof?

Answer 11

oh sorry
since the max of sine is 1,
\[|x\sin(x)-y\sin(y)|\leq 2|x-y|\] and so you can pick \(\delta=\frac{\epsilon}{2}\)

Answer 12

how did you get to x sin (x). Do you mean x sin (1/x)?
Also how did you get to 2|x-y|? Where did the 2 come from? I'm just trying to understand it.