prove lim, x->a, x^2 sin(x) = a^2 sin(a) using epsilon-delta

[arman_phsa]

please prove with e(epsilon)-d(delta) method

limit (x goto a) x^2 *sin(x) = a^2*sin(a)

thx

 

[stapel]

prove with e(epsilon)-d(delta) method

limit (x goto a) x^2 *sin(x) = a^2*sin(a)

What are your thoughts? What investigations have you done? What have you tried? How far did you get? Where are you stuck?

(As currently posted, it almost reads like you're expecting somebody to cheat and complete your assignment for you, which of course is not what you meant. But until you show what you've done so far, or specify that you're needing lessons on epsilon-delta proofs, then we really can't "see" where you're having difficulty.)

Please be complete. Thank you!

Eliz.

 

[PAULK]

please prove with e(epsilon)-d(delta) method

limit (x goto a) x^2 *sin(x) = a^2*sin(a)

thx

You want to prove

| x^2 sin x - a^2*sin(a) | < e

whenever | x - a | < d

So you want to make it as hard for yourself as possible and show you can still do it.

Take your expression:

| x^2 sin x - a^2*sin(a) |

and consider the worst cases for sin x and sin a. They could be as big as 1, right? So the worst (biggest value) of this would be:

| x^2 - a^2 |

We want to get (x - a) in there, don't we? Factor.

| (x + a)(x - a) |

Now how bad (big) could (x + a) be? Let's assume that if x is supposed to be close to 'a' it won't be further away than 1 unit. In that case:

a - 1 <= x <= a + 1

In that case the worst would be x = a + 1 and x + a = 2a + 1.
(Yes, a could be negative, but lets leave that for another day.)

Now we have:
| (2a + 1)(x - a) | =

| (2a + 1) | |(x - a) | =

That's as bad as it gets. Can we still make that < e?

Yes, and you can handle it from here, right?