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

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