Proving the Limit of x^2sin(1/x) as x Approaches 0 Using Delta-Epsilon Proof

[Math_Geek]

Homework Statement

limit as x goes to 0 of x^2 sin(1/x)=0

Homework Equations

Use delta-epsilon proof

The Attempt at a Solution

So |f(x)-L|=|x^2 sin(1/x)|=|x^2||sin(1/x)| and I know that sin(1/x) is bounded by one. I am not sure how to finish because of the x^2.

 

[tiny-tim]

… one step at a time … !

So |f(x)-L| = |x^2 sin(1/x)| = |x^2||sin(1/x)| and I know that sin(1/x) is bounded by one. I am not sure how to finish because of the x^2.

Hi Michelle!

I can't make out whether you've got the answer or not.

You must practise stating things clearly.

And thinking clearly! One step at a time!

First step: what do you think the limit is?

Second step: why do you think it's that (in layman's terms)?

Third step: put second step into delta-epsilon form.

… three steps to happiness! … ​

Have a go!

 

[Math_Geek]

the limit is 0, I think this because it given, also I know sin(1/x) is bounded by so that leaves x^2, so if I choose my epsilon to be sqrt epsilon. then when you square the x then x^2<epsilon
Right?

 

[tiny-tim]

… oh happiness … !

the limit is 0, I think this because it given, also I know sin(1/x) is bounded by so that leaves x^2, so if I choose my epsilon to be sqrt epsilon. then when you square the x then x^2<epsilon
Right?

Very good!

hmm … now we've got you thinking clearly, how about writing clearly?

You see … you meant "… my delta to be sqrt epsilon", didn't you?

… and it would be much better if you got into the habit of actually writing "Given any epsilon > 0, then for any x with |x| < √epsilon, |x^2sin(1/x) |≤ x^2 < epsilon; therefore lim(x¬0) = 0", instead of just thinking it!