11dude take log on both sides
then use lim x->0 ax-1/x=loga
u will get the answer
16 Answers
11
11dude.........
I already have the ans. & I know how to solve it.[1][1]
This is for others.
11yes the answer should be n
tke n common from the equation
andar waali term mein 1 bach jaaega after cutting shutting
so we r left with n
rkrish now disclose the answer
11[b]L = Lt x→0 ( 1cosec2x + 2cosec2x +...+ ncosec2x ) sin2x
Tr = r cosec2x
As x→0 , Tr (r=1,2,..,n) → ∞
Thus, as x→0 , AM → GM
L = Lt x→0 { n . ( 1cosec2x + 2cosec2x +...+ ncosec2x )/n } sin2x
= Lt x→0 { n . ( (1.2.3...n)(1/n) . cosec2x ) } sin2x
= Lt x→0 { n sin2x . ( (1.2.3...n)(1/n) . cosec2x . sin2x ) }
= Lt x→0 { n sin2x . (n ! ) (1/n) }
= n0 . (n ! ) (1/n)
= (n ! ) (1/n)
1u dont need to do all this...for u just need to take out n^cosec2x to get n((1/n)^coesc2x + (2/n)^cosec2x+...+1)^sin2x now this thing inside the bracket tends to 1 as x-> 0, leaving us with n as the answer..
1u dont need to do all this...for u just need to take out n^cosec2x to get n((1/n)^coesc2x + (2/n)^cosec2x+...+1)^sin2x now this thing inside the bracket tends to 1 as x-> 0, leaving us with n as the answer..
11" now this thing inside the bracket tends to 1 as x-> 0 "
how [7][7] (i may be asking a very foolish ques.)
11in that term within the brackets
except one all other terms are (<1)∞ so they all become zero
66rkrish
what you have done is not correct. For example, for n=100, it can be easily seen from the plot below that the limit required is 100 and not (100!)1/100 as you have got. Can you guess the reason why your work is incorrect?
(the above output is from Mathematica)
9AM GM equality is a very loose inequality
avoid it in limits
341Also there is no basis to say that as x→0, AM→GM. That will happen if each of the terms tend to each other. But 1cosec2x will tend to a different limit than 2cosec2x as x→0.
11sir ji meko prophet sir and anand sir ki bhavnaain samajh nahin aa rahi
isn't the answer n and wat's rong in #7[7]