# Delta-Epsilon Proof

• Aug 6th 2009, 03:34 AM
amee
Delta-Epsilon Proof
This was a problem from my calculus final a year ago that I've wanted to see for a while:

Use the delta-epsilon definition of a limit to prove that the limit as x approaches 0 of f(x) = sin(x)/(x^2 +1) is 0.

I tried using the squeeze theorem in an effort to bound sin(x), because I really don't know how to deal with sin(x) in a delta epsilon proof. This problem has just been on my mind for a while. Thanks for the help!
• Aug 6th 2009, 03:40 AM
CaptainBlack
Quote:

Originally Posted by amee
This was a problem from my calculus final a year ago that I've wanted to see for a while:

Use the delta-epsilon definition of a limit to prove that the limit as x approaches 0 of f(x) = sin(x)/(x^2 +1) is 0.

I tried using the squeeze theorem in an effort to bound sin(x), because I really don't know how to deal with sin(x) in a delta epsilon proof. This problem has just been on my mind for a while. Thanks for the help!

For all $x \in \mathbb{R}$ and $x\ne 0$: $|\sin(x)|< |x|$

CB
• Aug 6th 2009, 03:47 AM
amee
Can you show me how to do this proof? CB, what you wrote is true, but I'm not familiar with delta-epsilon proofs to know how to incorporate it into the rest of the proof.
• Aug 6th 2009, 03:59 AM
Calculus26
You want d such that |x- 0| < d then |f(x) - 0 | < e

|f(x) - 0 | = | sin(x)/(x^2+1)| < |sin(x)|

take d = arcsin(e)

then if |x| < arcsin(e) then |sin(x)| < |sin(arcsin(e)| = e
• Aug 6th 2009, 06:46 AM
Jester
Quote:

Originally Posted by Calculus26
You want d such that |x- 0| < d then |f(x) - 0 | < e

|f(x) - 0 | = | sin(x)/(x^2+1)| < |sin(x)|

take d = arcsin(e)

then if |x| < arcsin(e) then |sin(x)| < |sin(arcsin(e)| = e

But since $|\, \sin (x) \,| < |\, x \,|\; \text{for}\; |\, x \, | > 0$ then you can chose $\delta = \varepsilon$ .