Q.1 Find :   cos(x/2) cos(x/4) cos(x/8) ..........cos(x/2ⁿ)?

Q.2 Let f(x) =  sin²ⁿ (x).find the number of points    where f(x)is discontinuous?

is the answer for the first one 0?

Sorry, answer to the 1st ques is   .

Don' t worry let's have a try again.

but wont you apply the limit after getting sin x/x ?

and then tha answer is zero.

i think that it will be n->infinity then

cosx/2 cosx/4..............................cosx/2ⁿ=sin x/2ⁿsin x/2ⁿ (u can get this relation by induction or dividing n * by2 on lhs)

lim_n->infinitysin x/2ⁿsinx/2ⁿ (div n mul by x/2ⁿ sinx/2ⁿ /x/2ⁿ becomes 1 and 2ⁿ will be cancelled out and we get the ans sinx/x)

pls tell me if im wrong

I don' t know the solution .I knew only the answer. This ques is from brilliant tutorial study material.

let someone else try.

@ studen9t_iit

u should post the solution so that others can check whether u r right or wrong or may be the ans is wrong.

@nl

I could not get what u r trying to do. please do properly.

i m saying that instead of having x  tends to infinity it will be n tends to infinity otherwise the ans can't be sinx/x because if x tends to infinity  then the ans can't be in terms of X

I REQUEST PLS CHECK THE QUES AGAIN 

Sorry that was my mistake.

In ques there is    instead of

@nl

please do ur solution clearly, may be u r right.

 pls tell me in which step u r facing problemi.e the solution earlier posted by me.

i m not able to understand both the steps written by u specially the 1st one.

2nd one is really easy.

The critical points are the ones where sin x = 1 or sin x = -1. Here the limit is 1

for -1< sin x< 1 the limit is 0. So the points of discontinuity are:

hey yeah i did think it should be n tendiing to infinity..

but yeah whatever nl has done is correct...