3i also have the same dbt asish.i am waiting for a reply.
lim x→0 (1cosec2x+2cosec2x+3cosec2x+....Ncosec2x)sin2x=?????
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19 Answers
24@ bhatt sir
didnt get the first step of ur solution [2][2]
can u explain it ???
106manipal: 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-->∞ ??
11@ashish
lim(x→0) (1cosec2x+2cosec2x+...+ncosec2x)sin2x
= ncosec2xsin2x[(1/n)cosec2x+(2/n)cosec2x+....+1]sin2x
= n[(1/n)cosec2x+(2/n)cosec2x+....+1]sin2x
= n (as r/n < 1 and hence (r/n)∞→0)(further as n does not tend to ∞, so we dont have (→0)infinite no. of times. so all the terms can be assumed to be zero)
how could u divide by ncosec2x when x->0
106nopes... see (N-1)/N < 1
(any number less than 1)∞ →0 isnt it?
1arent we complicating this problem?
i mean as asish say, take out Ncosec2x out of the bracket, to have
N*(1+(N-1/N)^cosec2x+.....)sin2x
=N rite.
106manipal: ur approach is wrong
logy=sin2x log(N!)cosec2x yahi wrong hai.
lim(x→0) (1cosec2x+2cosec2x+...+ncosec2x)sin2x
= ncosec2xsin2x[(1/n)cosec2x+(2/n)cosec2x+....+1]sin2x
= n[(1/n)cosec2x+(2/n)cosec2x+....+1]sin2x
= n (as r/n < 1 and hence (r/n)∞→0)(further as n does not tend to ∞, so we dont have (→0)infinite no. of times. so all the terms can be assumed to be zero)
11PLEASE DUNNO RELY ON THIS ONE AND DO CHECK FOR CONFORMATION[3]
lim x→0 (1cosec2x+2cosec2x+3cosec2x+....Ncosec2x)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]
3N [ (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 ...
3ok sir thanx for pointing the mistake.ive deleted that post to prevent further confusion.
341Sandwich 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.
62I dont think that is perfectly correct..
because it is of the form zero to the power zero..
that is not necessarily zero...
3I DUNNO THE ANS AND ALSO WE CANT DO LIKE THIS SOMETIMES IT LEAD TO US A WRONG ANS.