Question

limit question

Answer 1

find the limit

\(\large \lim \limits_{n \to \infty} 2^n \sin (2^{1-n} )\)

Answer 2

wolfram is giving 2.. any ideas on solution.. ?
http://www.wolframalpha.com/input/?i=+lim+n+to+infty+%28+2%5En+*+sin+%282%5E%7B1-n%7D+%29%29

Answer 3

I graphed it on my calculator and I am also getting 2. I'm not clear why though . . .

Answer 4

yah it doesnt look obvious... i got stuck at this limit while doing below problem :
http://openstudy.com/study#/updates/530720dce4b00fb1e2378075

Answer 5

try l-hospital rule

Answer 6

i guess i can try... but i need to change it to indeterminate form first ? can u help wid that ? :)

Answer 7

I am not sure why you would get your question from the question of shamil, since I would obtain \(\displaystyle\large\lim_{n\rightarrow\infty}\frac1{2^{n+1}}\sin2\).

Answer 8

thats from shamil's question ?

Answer 9

actually i got till below and got stuck...
\(\large \frac{\sin 2}{\lim \limits_{n \to \infty} 2^n \sin (2^{1-n} ) }\)

Answer 10

Oh, I used the same approach as you did, but look at my result obtained...

Answer 11

how u got rid of sin in the bottom ?

Answer 12

But I don't think you can move the limit to the denominator? e.g. \(\displaystyle\lim_{n\rightarrow\infty}\frac1n\ne\frac1{\lim\limits_{n\rightarrow\infty}n}\)

Answer 13

Sorry, I forgot to divide the whole thing by sin(1/2^n) haha

Answer 14

okay

Answer 15

lets see how to go from here :)

Answer 16

i feel changing it to indeterminate form may give somthing... but not getting ideas how to turn it..

Answer 17

ganeshi...
cant we complete this by using a substitution... ?

Answer 18

anything is okay... .

Answer 19

\[\frac{ 1 }{ 2^{n} } = t\]

Answer 20

brilliant !!!

Answer 21

when\[n \rightarrow \infty \]
\[t \rightarrow 0\]

Answer 22

yes ! that gives indeterminate form xD

Answer 23

now u can write
\[\lim_{t \rightarrow 0}\sin \frac{ 2t }{ t }\]

Answer 24

next l hospitals... sweet :)

Answer 25

yep :)

Answer 26

@ganeshie8 see the original question

Answer 27

thanks both :)))

Answer 28

OMG !! It 's really feels GRT ... I was able to help the person I admires most in OS... i m going to die :)

Answer 29

I wish to you a fruitful life in both OS and your real life :D

Answer 30

you helped a lot really... been stuck on this like forever...
i can sleep peacefuly tonight xDD ty again :))

Answer 31

u. r welcome ...and tnx kennylau.. :)

Answer 32

i dont know if this is same as ***[ISURU]*** approach
\(\large \lim_{n\to \infty }\dfrac{2\sin(2^{1-n})}{2^{1-n}}\)
let \(2^{1-n}=\dfrac{2}{2^n}=t\to 0\)

\[\lim_{t\to 0}\frac{2\sin t}{t}=1\]