evaluate

I Am Not Sure of the Answer

let x=1/t

then limit changes to lim t=>0 sec^-1 (1/(1+t))

now let  t=cos(theta)

then sec^-1 (1/1+cos(theta))

==>> sec^-1 (1/2cos^2(theta/2))

==>> sec^-1 (sec^2(theta/2)/2)

==>> [((theta/2)/2)*sec (theta/2)]/2

now (theta)=(cos^-1 (t))/2

==>>{[ cos^-1 (t)/2 ] *1/[cos(cos^-1 (t)]}/2

==>>{ (cos^-1(t)/2) *1/[t/2] }/2

now put t=0 you will get Zero

You can also try with LN Hospital rule its very easy with it

rate me if found useful

if not understand i can give you better explation then this

I think question is incorrect. For positive values of x, numerator is less than denominator ( i.e it is a fraction) for which function is not defined

arre yaar bracket se x bahar nikal lene par sec^-(1/1+1/x) bachta hai. since x tends to infinity we get sec^-1 i.e 0 is the answer

Thanks for clearing y doubt

my dear friends

option for given question are

0

pie

pie/2

none

and correct answer is none

it seems ananth910 is correct but has failed to give satisfactory explanation

and my dear friend oneyeartogo the question is absolutely right

 

yes i m also getting the ans as 0

tell me if i m correct

also @oneyeartogo when x tends to infinity x tends to x+1 so we get sec inverse 1 which is 0.

but i still think that ur doubt is genuine and a good thought provoking one .I might be wrong but i have given my view on this.

hope u got it...

@animal you are absolutely right.

I thought the same way. As x ----->infinity the function just comes into existense or as you put it x ----> x+1

@thedumbheadwithnobrain          

I don't think your solution is right. Your doubt is same as mine (answered above)

Look at line number 9