Question

Limit

Answer 1

\[\lim_{n \rightarrow \infty} [(1+1/ n^2)(1+2^2/n^2)......(1+n^2/n^2)]^{1/n}\]

Answer 2

why dont u try taking log..

Answer 3

i dont know how to do it

Answer 4

if we simply put n=infinity result comes out to be \[1^{0}\]

Answer 5

is it an indeterminate form?

Answer 6

take log on both sides
ln L = 1/n x sigma ln (1 + r^2/n^2)

Answer 7

take r/n = x, 1/n= dx

Answer 8

hence
ln L = ln(1 + x^2)dx from 0 to 1

Answer 9

now use integral by parts to integrate this

Answer 10

is it an indeterminate form?
1^0

Answer 11

i told u..did u get it??????????????????????????/

Answer 12

deeprony? r u getting it?

Answer 13

i didnt understand what you said

Answer 14

see ill tell u the general method
whenever u hav
\[\sum_{0}^{\infty}f(r/n) \times 1/n\]

Answer 15

yes.. what do i do?

Answer 16

then put r/n = x and 1/n = dx

Answer 17

so u hav \[\int\limits_{0}^{1}f(x)dx\]

Answer 18

samjhe??

Answer 19

yea but why did u set the upper limit to 1??

Answer 20

when r=n then r/n is 1 isnt it/?

Answer 21

oh yes.. i get it now...

Answer 22

\[given=\lim_{n \rightarrow \infty} e^{\frac{\ln (1+\frac{1}{n^2})+...+\ln(1+\frac{n^2}{n^2})}{n}}\]
apply l hospitals rule.

Answer 23

please i am having a doubt
if we directly put n=infinity then the limit is coming out to be 2^0
is this an inderminate form???

Answer 24

its coming infinity^0

Answer 25

y so saubhik n^2 is in the denominator
as n tends to infinity 1/n^2 , 2/n^2 ,... becomes 0

Answer 26

see the last term its 1+n^2/n^2.
as n -> infty then thus term tends to infty.
understood?

Answer 27

cancel the n^2
last term comes out to be 2

Answer 28

wow wow man!!! Mankind has not been able to understand till today what is infty/infty and yet u claim its 1!

Answer 29

ok
my bad
i should have been more careful

Answer 30

hey him r u der?

Answer 31

get it?

Answer 32

yea.. it worked out..=)

Answer 33

chek dis one out
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4dd66882d95c8b0b951d5ec4