IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

HOTGUY

i got few problems in limit

Q1>

_{[ x]}

_{[ 0]} [nsix/x] where n belongs to natural no.??? {[]=gif}

Q2>

_{[x ]}

_{[0 ]} [ntanx/x] where n belongs to natural no.??? {[]=gif}

johri_anshuman

in both the problems the limit does not exist

_{[ x]}

_[0+] [nsinx/x] =n-1 as sin(x)/x<1 for x>0

_{[ x]}

_[0-] [nsinx/x] =n as sin(x)/x>1 for x<0

similarly

_{[ x]}

_[0+] [ntanx/x] =n as tan(x)/x>1 for x>0

_{[ x]}

_[0-] [ntanx/x] =n-1 as tan(x)/x<1 for x<0

so in both the questions limit does not exist.

bumba

ia the answer n-1 in both the cases

spideyunlimited

yes n - 1 in both cases

asmit

The important inequality to use here is tanx>x>sinx.So that would mean sinx/x<1 and tanx/x>1 .In the limiting case when x is close to 0,the 2 ratios get very close to 1.However they are never equal to 1 and maintain the respective inequalities mentioned.Hence in the first case the quantity in box is slightly less than n and hence answer is n-1.On the other hand the term in the 2nd function inside the box is slightly greater than n and hence answer is n.Another thing to note here is that since tanx/x and sinx/x remain positive on the left hand side of 0.So taking limit from that side too would yield the same answer.However the initial inequality used is valid between 0 and pi/2

HOTGUY

ok i accept ur answer asmit,

still i have few conceptual doubts

now please tell me

Is

_{[ x]}

_{[y ]} ^{[2 ]}

x²+x+c = ^{[ 2]}

_{[ x]}

_{[ y]} x²+x+c ???? why(or any function residing any expression)

if so then

cant we apply limit inside G.I.F and since we know that

_{[ x]}

_{[0 ]} sinx/x = 1

and _{[ x][0 ]} tanx/x = 1

then the value will depend upon the nature of n?

asmit

See limit can be taken inside f of f(g(x)) as long as it exists.I too did the same thing in my soln.I took the limit inside the box but I also mentioned the inequality.So in the limiting case we may write the 2 ratios as 1 but actually they are slightly less or more than 1.Hence the result.See that is where the inequality has to be applied because x tends to zero,not equal to it,so the inequality is strictly valid and therefore the result.

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