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jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0plz explain too . i found out it to be 0. u r right it is 1 how

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1answer is 1,actually it's proof is a bit theoretical .if you want any proof please refer some standard text

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1it's proof is well known and is known as sandwich theorem

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0as x reduces, the sin of x also reduces, then why 1

ankit709
 2 years ago
Best ResponseYou've already chosen the best response.0if u put the value of limit then u will get 0/0 form

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1actually 0/0 is a indeterminate value

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0derivative ? but i don't know the derivative i'm beginner but soon be older :D :D hahahaa

ankit709
 2 years ago
Best ResponseYou've already chosen the best response.0this is L hospital rule

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0furan.. yah that would be indeterminate state

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0ohhh.. hospital ....

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1I think the proof starts from some sort of triangle & using some inequality conditions

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1buddy L'hospital rule is not the actual prrof

ankit709
 2 years ago
Best ResponseYou've already chosen the best response.0I KNOW but u also know that proof of this equation is 2 long

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0i 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 ...

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0really too long ? i thought it would be very simple

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1okay then you better use L'hospital's rule then

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0yah.... i will ... :) thanks all frnds in here .. :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1actually sandwich theorem is like the foundation of calculus

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1proof is not that long but concept is difficult

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0sandwich.... i need to work up at great deal :) :) i will .. bcoz i want to be so perfect in calculus :)

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1I remembers the last step

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0it's alright no problem. thanks you all alot...... really helpful for me.... :)

CeW
 2 years ago
Best ResponseYou've already chosen the best response.0sin(x) for small x is x. x/x is 1 for any x.

furan143
 2 years ago
Best ResponseYou've already chosen the best response.1cosx<(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

CeW
 2 years ago
Best ResponseYou've already chosen the best response.0sin(x) / x does not approach 0 in the limit of x > 0 !!

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0santosh r u there?

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0hi experimentX, how are you brooo...

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0doing fine ... i might be offline pretty soon .. dw:1334192708048:dw

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0before being, could you expect to have some time with me or not ?

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0open these three :)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0you 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

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0i want to know why epsilon is smaller than a  x0 and x1  a

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0yah.... both approach toward 0 i saw that

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0It's not both approach to zero ... it's how their ratio approach as Q approaches to zero > that is limit.

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0why epsilon is smaller than a  x0 and x1  a ............................... :)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0how is epsilon defined??

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0i know how epsilon is defined but why epsilon is smaller than a  x0 and x1  a ............................... :)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0i don't know how epsilon is defined ... is it xa < epsilon or f(x)  L  < epsilon

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0omggggggggggggg... how i can get answer to my question... here no one knows about limit definition properly..... ... leave it i try to google it... :D

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0thanks experimentX..... :) :)

jatinbansalhot
 2 years ago
Best ResponseYou've already chosen the best response.0ExperimentX... I go and study mathematics more and more if again any ques i would come back :) take care brooo :)
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