Expert: Paul Klarreich - 10/4/2009

Question
QUESTION: plz solve this for me

Using the formal epsilon − delta definition of derivative, show that
lim xapprochaes 0 x^2sin(x)=0

ANSWER: Questioner: waqas
Country: Australia
Category: Calculus
Private: No
Subject: Epsilon Delta problems
Question: plz solve this for me

Using the formal epsilon − delta definition of derivative, show that
lim xapprochaes 0 x^2sin(x)=0
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Hi, Wagas,

You are not using the vocabulary correctly.  You should either write:

Using the formal epsilon − delta definition of limit

or

Using the formal definition of derivative as a limit.

Misusing (and therefore misunderstanding) the vocabulary is a good way to fail mathematics.
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I shall assume you mean:

Using the formal epsilon − delta definition of limit

show that lim [x -> 0] x^2sin(x)=0

You want to show that for any  e > 0 [e is epsilon, d is delta]

you can find  d > 0 such that whenever  |x - 0| < d,

| x^2sin(x) - 0 | < e

Start with

| x^2sin(x) - 0 | = | x^2sin(x) |

| x^2sin(x) | = | x^2 | | sin(x) |

| x^2 | | sin(x) | <= | x^2 | (1) = | x^2 | = x^2, since x^2 is never negative.

Now we want  x^2 < e, so take  |x| < sqrt(e).

That is your delta.

You can now write the proof:

| x^2sin(x) - 0 | < | x^2sin(x) | =

| x^2||sin(x) | <= |x^2|

= |x|^2 < | sqrt(e) |^2 = e

Done


---------- FOLLOW-UP ----------

QUESTION: i did not understand after that "u can now write the proof" i mean how this proofs that  x^2sinx=0

Answer
QUESTION: i did not understand after that "u can now write the proof" i mean how this proofs that  x^2sinx=0
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I did not say 'u can..', please.  
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And what followed IS the proof.
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It is the proof that LIM(X -> 0) x^2sinx = 0

I'll do it over:

We have decided to set  

|x| < sqrt(e)

then

|x|^2 < (sqrt(e))^2

|x|^2 < e

|x^2| < e

And  

|sin x| < 1

So take both of those:

|x^2| < e
|sin x| < 1

and multiply:

|x^2 sin x| < e

Pretend to subtract 0:

|x^2 sin x - 0| < e

That is it;  whenever  |x - 0| < d = sqrt(e)

your |f(x) - 0 | < e.

That is the best I can do.