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

Mathematics
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=1
plz explain too . i found out it to be 0. u r right it is 1 how
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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Other answers:

it's proof is well known and is known as sandwich theorem
as x reduces, the sin of x also reduces, then why 1
if u put the value of limit then u will get 0/0 form
i will get 0
offcourse buddy
u can apply dy/dx
actually 0/0 is a indeterminate value
derivative ? but i don't know the derivative i'm beginner but soon be older :D :D hahahaa
this is L hospital rule
furan.. yah that would be indeterminate state
ohhh.. hospital ....
I think the proof starts from some sort of triangle & using some inequality conditions
buddy L'hospital rule is not the actual prrof
srry proof
I KNOW but u also know that proof of this equation is 2 long
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 ...
really too long ? i thought it would be very simple
okay then you better use L'hospital's rule then
yah.... i will ... :) thanks all frnds in here .. :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)
actually sandwich theorem is like the foundation of calculus
proof is not that long but concept is difficult
sandwich.... i need to work up at great deal :) :) i will .. bcoz i want to be so perfect in calculus :)
I remembers the last step
it's alright no problem. thanks you all alot...... really helpful for me.... :)
sin(x) for small x is x. x/x is 1 for any x.
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
sin(x) / x does not approach 0 in the limit of x -> 0 !!
you are right cew
santosh r u there?
yeah
hi experimentX, how are you brooo...
doing fine ... i might be offline pretty soon .. |dw:1334192708048:dw|
before being, could you expect to have some time with me or not ?
open these three :)
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
i want to know why epsilon is smaller than a - x0 and x1 - a
yah.... both approach toward 0 i saw that
It's not both approach to zero ... it's how their ratio approach as Q approaches to zero -> that is limit.
why epsilon is smaller than a - x0 and x1 - a ............................... :)
how is epsilon defined??
i know how epsilon is defined but why epsilon is smaller than a - x0 and x1 - a ............................... :)
i don't know how epsilon is defined ... is it |x-a| < epsilon or |f(x) - L | < epsilon
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
thanks experimentX..... :) :)
ExperimentX... I go and study mathematics more and more if again any ques i would come back :) take care brooo :)
sure ...

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