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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
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 : f(x)=(x−1)(x−1)(x) , now on a day only one toy is made by labours , now manager wants how much profit he got ? i,e :f(1)=? now what did labours say ? Now mathematicians needed to invent Limit's concept! So by using it they got : f(1+)→1f(1−)→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:
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
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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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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i too think so..!
No confusion, 1st question answer is 0 and second one's answer is 1
i think both are zero, whats ur answer.
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Yup both are zero ....
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.....................
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.
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 ............. ;)
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 : f(x)=(x−1)(x−1)(x) , now on a day only one toy is made by labours , now manager wants how much profit he got ? i,e :f(1)=? now what did labours say ? Now mathematicians needed to invent Limit's concept! So by using it they got : f(1+)→1f(1−)→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:
1)RHL=x→0+lim⌊xsinx⌋=x→0+lim⌊0+sin0+⌋=x→0+lim⌊1−⌋=0LHL=x→0−lim⌊xsinx⌋=x→0−lim⌊0−sin0−⌋=x→0−lim⌊1−⌋=0
2)x→0limxsinx=1⇒⌊x→0limxsinx⌋=⌊1⌋=1∵0+sin0+→1−&0−sin0−→1−
Conclusion:: x→0limxsinx=1x→0limxsinx=1−
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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 ?
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
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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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
you are totally wrong