find the limit

@ bhatt sir
didnt get the first step of ur solution [2][2]
can u explain it ???

manipal: see this
lim(x-->0) (sinx+2x)/x
How can u divide by x (x-->0) and say that it is 1+2=3??

same way here. ive just divided by an expression
or see this
lim(x→∞) x²+2x2x²+1 = lim(x→∞) 1+2/x2+1/x² = 1/2
How can we divide by x² as x-->∞ ??

nopes... see (N-1)/N < 1
(any number less than 1)^∞ →0 isnt it?

arent we complicating this problem?
i mean as asish say, take out N^cosec2x out of the bracket, to have
N*(1+(N-1/N)^cosec²x+.....)^sin2x

=N rite.

PLEASE DUNNO RELY ON THIS ONE AND DO CHECK FOR CONFORMATION[3]

lim x→0 (1^cosec2x+2^cosec2x+3^cosec2x+....N^cosec2x)^sin2x=y

then

logy=sin²x log(N!)^cosec2x
log(logy)=log(logN!)

y=N!

This idea came randomly in my mind
please excuse me if i am wrong and sum1 pls solve this 1
This question is pondering me now[2]

okk......that was silly doubt[3][3]

thanxxx sir[1][1]

Each of the summands (except the first) is less than 1.

N [ (1/N)^COSEC^2 X + 2/ ............................+1]^SIN^2 X .....!!!!!!!!!!!!!!!!!!!!!!!!!!

we know ...1/n is less than 1 and ... cosec thingi tends to infinity as x -->0 !!!!!!

so all those become 0 .......... except the 1 ..... so we get 1 *N =N ......

SRY IF RONG ...

Sandwich Principle can easily settle this:

We have

1< 1^{\csc^2 x} + \left(\frac{1}{N} \right)^ {\csc^2 x} + \left(\frac{2}{N} \right)^ {\csc^2 x}+...+ \left(\frac{N-1}{N} \right)^ {\csc^2 x}< N

so that

1^{\sin^2 x}< \left(1^{\csc^2 x} + \left(\frac{1}{N} \right)^ {\csc^2 x} + ...+ \left(\frac{N-1}{N} \right)^ {\csc^2 x} \right)^{\sin^2 x} < N^{\sin^2 x}

Now take limits as x→0

Note: Here RHS bound we have taken is pretty weak, but it works just fine.

But, we can improve the bounds by saying

1^{\csc^2 x} + \left(\frac{1}{N} \right)^ {\csc^2 x} + ...+ \left(\frac{N-1}{N} \right)^ {\csc^2 x}< \frac{1}{N} + \frac{2}{N}+...+\frac{N}{N}

or further

1^{\csc^2 x} + \left(\frac{1}{N} \right)^ {\csc^2 x} + ...+ \left(\frac{N-1}{N} \right)^ {\csc^2 x}< \frac{1}{N^2} \left(1^2+2^2 + ...+N^2 \right)

and so on.

We get RHS<(N+1)/2 in the 1st case which is a better bound than N

When we choose squares, we get N(N+1)(2N+1)/6N² which is a further improvement over (N+1)/2.

This could prove useful somewhere.

my ans. matches wit bipins...........

pls try it

I DUNNO THE ANS AND ALSO WE CANT DO LIKE THIS SOMETIMES IT LEAD TO US A WRONG ANS.