what is

1) \(\displaystyle \lim_{x\to 0}[\frac{sinx}{x}]\)

and

2) \(\displaystyle [\lim_{x\to 0} \frac{sinx}{x}]\)

where \([.]\) is greatest integer function.

what is

1) \(\displaystyle \lim_{x\to 0}[\frac{sinx}{x}]\)

and

2) \(\displaystyle [\lim_{x\to 0} \frac{sinx}{x}]\)

where \([.]\) is greatest integer function.

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TopNewestNo confusion, 1st question answer is 0 and second one's answer is 1

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i too think so..!

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I was always taught that lim x --> 0 sin(x)/x --> 1 By looking at the series expansion for sin(x) , dividing by x, we end up with a convergent series that starts with one giving the limit of 1! or you can use the squeeze theorem as is done here; https://www.khanacademy.org/math/ap-calculus-ab/limits-from-equations-ab/squeeze-theorem-ab/v/proof-lim-sin-x-x I think this is done several places in your questions and needs to be corrected!!

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Answers are 0 and 1. First is a very standard question , its obvious to see because sin(x)/x is just less than one as x tend towards zero. For the second part , the limit evaluates to 1 , hence its GIF is 1

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these are not confusions , these are questions , So better thing is Please mention your doubts , so that may be anyone help you .! Without knowing your doubt , all others are helpless .

But Okay , as far I comprehended , I think You need to recall very elementary definition , Which states that (Please don't be angry as I'am saying this in my own language , And forgave me if i do some typo :P)

'' while we evaluating and limit and we find condition as LHL=RHL=finite , then we say limit exist , right ....? But what would you say about that value , I mean will you say It is exact or an approx ....? ''

Answer is , that calculated value is Exact , not limiting! now you say why ....? before answering this ask yourself that what is our aim , for calculating it's limiting value ? Yes , for explaining this , consider an example: Let an factory produces some toys , and number of toys it produces per day is 'x' , and somehow , It is found that (by analysing labour strength , needs of products etc. ) that profit per day function is expressed as : \(\displaystyle{f\left( x \right) =\cfrac { (x-1)(x) }{ (x-1) } }\) , now on a day only one toy is made by labours , now manager wants how much profit he got ? i,e \(\quad :\quad f(1)=?\) now what did labours say ? Now mathematicians needed to invent Limit's concept! So by using it they got : \(f\left( { 1 }^{ + } \right) \rightarrow 1\\ f\left( { 1 }^{ - } \right) \rightarrow 1\)

So " now what does labours say to manager ? does they say that profit is 1+ (means little more than one rupees) ? or does they say profit is 1- (means little less than one rupees )?

Hence for avoiding such complications and ambiguty , we (mathamaticiens) decided that "

Limiting value is always exact"And Think yourself , we want to calculate limiting value of something and what we got ? what would we say .... that take limit of answer also after taking limit of question.!

Hence remember that " Limit of functions is always exact". So in current situation:

\(\displaystyle{1)RHL=\lim _{ x\rightarrow { 0 }^{ + } }{ \left\lfloor \cfrac { \sin { x } }{ x } \right\rfloor } =\lim _{ x\rightarrow { 0 }^{ + } }{ \left\lfloor \cfrac { \sin { { 0 }^{ + } } }{ { 0 }^{ + } } \right\rfloor } =\lim _{ x\rightarrow { 0 }^{ + } }{ \left\lfloor { 1 }^{ - } \right\rfloor } =0\quad \\ LHL=\lim _{ x\rightarrow { 0 }^{ - } }{ \left\lfloor \cfrac { \sin { x } }{ x } \right\rfloor } =\lim _{ x\rightarrow { 0 }^{ - } }{ \left\lfloor \cfrac { \sin { { 0 }^{ - } } }{ { 0 }^{ - } } \right\rfloor } =\lim _{ x\rightarrow { 0 }^{ - } }{ \left\lfloor { 1 }^{ - } \right\rfloor } =0}\)

\(\displaystyle{2)\lim _{ x\rightarrow 0 }{ \cfrac { \sin { x } }{ x } } =1\quad \Rightarrow \left\lfloor \lim _{ x\rightarrow 0 }{ \cfrac { \sin { x } }{ x } } \right\rfloor =\left\lfloor 1 \right\rfloor =1\\ \\ \because \cfrac { \sin { { 0 }^{ + } } }{ { 0 }^{ + } } \rightarrow { 1 }^{ - }\\ \& \cfrac { \sin { { 0 }^{ - } } }{ { 0 }^{ - } } \rightarrow { 1 }^{ - }}\)

Conclusion:: \[\displaystyle{\lim _{ x\rightarrow 0 }{ \cfrac { \sin { x } }{ x } } =1\quad \\ \lim _{ x\rightarrow 0 }{ \cfrac { \sin { x } }{ x } } \neq { 1 }^{ - }}\]Log in to reply

they are greatest integer functions not least integer functions bro

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So what ? Yes , he also type as greatest integer function , what are you trying to say ?

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how many of u r giving JEE MAIN?????????

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a lot many,

ronak, deepanshu, pratik, saketh, mahimn, pranjal, azhaghu, shashwat, sheshansh, krishna, kishlaya, k.shekhawat, vraj, abhishek, harshvardhan, you and .......

me too

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all d best to everyone ............. ;)

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1 st is zero.2nd is 1..i am 99% sure

bcosz sinx<x as x goes from 0 to 1..whih implies sinx /x <1

thus [ sinx/x ] =0 as x tend to zero

2nd case is simple [1]=1

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the answer i have says that both are zero as when x approaches zero sinx/x has value slightly less than 1

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you are right that sinx/x is slightly less than 1,

but limit of something (if it exists), is always a fixed value and not a tending value.

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actually the first one depends on which side of zero r u approaching............. for positive side or negative while the second one is independent of this

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thsnk u for ur suggestion and i guess the conversation is over , punk

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thanks for ur that word called 'punk' :)

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i mean to say student on brilliant XD.

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so the first one can be, what 0 or something else

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both 0 here but the difference may be noted in some other cases.....................

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i think both are zero, whats ur answer.

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Yup both are zero ....

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answer to 1 : 0

answer to 2 : 1

@Tanishq Varshney , 100% sure

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no @Karan Siwach it can never be greater than 1 go by expansion of sinx .......... ans is same @Tanishq Varshney ... but u will notice the difference in same problem of tanx/x.......... just go by expansion

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limit of something (if it exists), is always a fixed value and not a tending value

now, again go through 2) lim x->0 sinx/x = 1 and not tending to 1

@Madhukar Thalore

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Comment deleted Mar 25, 2015

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you are totally wrong

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