Lim (x-sinx/x) sin(1/x)
When x-----0 ...

Question

Lim (x-sinx/x) sin(1/x)
When x----->0 ...

samigupta8 · Answer

The division sign is under the whole term x-sinx and not just sinx

anonymous · Answer

is it this question!

samigupta8 · Answer

@faiqraees

samigupta8 · Answer

Nope!...
It is sin(1/x) this term is in multiplication with first one..
And even x tends to 0

hartnn · Answer

you're stuck only fir sin(1/x) term right?

what is the range of sine function?

samigupta8 · Answer

Exactly! 
(-1,1)

samigupta8 · Answer

Oops...closed bracket i meant...

hartnn · Answer

ok,
so whatever number x tends to
sin (1/x) will ALWAYS be withing that finite range.
agree?

hartnn · Answer

within*

samigupta8 · Answer

It will be oscillating at an indeed rapid pace between the values -1,1

anonymous · Answer

it doesn't exist. and it diverges.

hartnn · Answer

right, we'll take your words :)

what is this product now:
0*(number oscillating at an indeed rapid pace between the values -1,1) =  ...?

samigupta8 · Answer

Bt this is not exact 0...the term u multiplied it with

hartnn · Answer

the 2nd term?
its a finite number right?

samigupta8 · Answer

Nope...the first one

hartnn · Answer

what?

1- sin x/x = 1-1 = 0 
basics of limits :P

anonymous · Answer

is it this!

samigupta8 · Answer

Sinx/x <1

anonymous · Answer

can you write the actual question which make sense .. put bracket to distinguish!

hartnn · Answer

ok, let me put it this way:
$\lim \limits_{x \to 0}\dfrac{\sin x}{x}=1$
agree?

samigupta8 · Answer

Yep...

samigupta8 · Answer

Do we need to separate limits like this 
Lim (f(x)*g(x)) can be written as 
Lim (f(x)) *lim(g(x))

hartnn · Answer

exactly

samigupta8 · Answer

Ok fine then...it"s done...

samigupta8 · Answer

Thanks...

zarkon · Answer

you need to use the squeeze theorem

samigupta8 · Answer

Thanks sir...! Bt it's an mcq type question