find out the limit of the following
( sin x ) / x where x -> 0

- anonymous

find out the limit of the following
( sin x ) / x where x -> 0

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- anonymous

=1

- anonymous

plz explain too . i found out it to be 0. u r right it is 1 how

- anonymous

answer is 1,actually it's proof is a bit theoretical .if you want any proof please refer some standard text

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## More answers

- anonymous

it's proof is well known and is known as sandwich theorem

- anonymous

as x reduces, the sin of x also reduces, then why 1

- anonymous

if u put the value of limit then u will get 0/0 form

- anonymous

i will get 0

- anonymous

offcourse buddy

- anonymous

u can apply dy/dx

- anonymous

actually 0/0 is a indeterminate value

- anonymous

derivative ? but i don't know the derivative i'm beginner but soon be older :D :D hahahaa

- anonymous

this is L hospital rule

- anonymous

furan.. yah that would be indeterminate state

- anonymous

ohhh.. hospital ....

- anonymous

I think the proof starts from some sort of triangle & using some inequality conditions

- anonymous

buddy L'hospital rule is not the actual prrof

- anonymous

srry proof

- anonymous

I KNOW but u also know that proof of this equation is 2 long

- anonymous

i think for the time being, it is enough that the limit of that is = 1 .... as i will advance in this calculus, i will come across the proof soon and become familiar with that ...

- anonymous

really too long ? i thought it would be very simple

- anonymous

okay then you better use L'hospital's rule then

- anonymous

yah.... i will ... :) thanks all frnds in here .. :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)

- anonymous

actually sandwich theorem is like the foundation of calculus

- anonymous

proof is not that long but concept is difficult

- anonymous

sandwich.... i need to work up at great deal :) :) i will .. bcoz i want to be so perfect in calculus :)

- anonymous

I remembers the last step

- anonymous

it's alright no problem. thanks you all alot...... really helpful for me.... :)

- anonymous

sin(x) for small x is x. x/x is 1 for any x.

- anonymous

cosx<(sinx/x)<1
here (sinx/x) is sandwiched between cosx & 1 (here not that cosx is also 1 when x tends to zero so sinx/x has no other option other than to tend to zero)
that's why it has got the name sandwich or squeeze theorem

- anonymous

sin(x) / x does not approach 0 in the limit of x -> 0 !!

- anonymous

you are right cew

- anonymous

santosh r u there?

- experimentX

yeah

- anonymous

hi experimentX, how are you brooo...

- experimentX

doing fine ... i might be offline pretty soon ..
|dw:1334192708048:dw|

- anonymous

before being, could you expect to have some time with me or not ?

- anonymous

open these three :)

##### 3 Attachments

- experimentX

you see arc there = it's value = \(r\theta\), when \(
\theta->0\), the ARC and H seem to be one line.
or h/rQ -> tends to one as Q->0
h/r = sinQ => sinQ/Q->1 as Q->0

- anonymous

i want to know why epsilon is smaller than a - x0 and x1 - a

- anonymous

yah.... both approach toward 0 i saw that

- experimentX

It's not both approach to zero ... it's how their ratio approach as Q approaches to zero -> that is limit.

- anonymous

why epsilon is smaller than a - x0 and x1 - a ............................... :)

- experimentX

how is epsilon defined??

- anonymous

i know how epsilon is defined but why epsilon is smaller than a - x0 and x1 - a ............................... :)

- experimentX

i don't know how epsilon is defined ... is it |x-a| < epsilon or |f(x) - L | < epsilon

- anonymous

omggggggggggggg... how i can get answer to my question... here no one knows about limit definition properly..... ... leave it i try to google it... :D

- anonymous

thanks experimentX..... :) :)

- anonymous

ExperimentX... I go and study mathematics more and more if again any ques i would come back :) take care brooo :)

- experimentX

sure ...

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