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Another limit question...

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\[\lim_{n \rightarrow \infty}\ln\left( 1+ \frac{ 4-\sin(x) }{ n } \right)^n\]
My approach was that I raised it to the power of e and then I Just found the limit of the stuff inside. But that's as far as I got.... I see a pattern of (1+(1/x))^n which is the definition of e but I don't really know about this one...
4 -sin x or 4 - sin n ?? if its 4- sin x, then its a constant...

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Nope that is not a typo. It is indeed a constant.
cool! so you have the function of the form \((1+ax)^{1/x}\) right?
How so?
Okay well I can see it if you make a substitution.
Go on.
good, i'll tell you what we do in case of \((1+ax)^{1/x}\) -- > \((1+ax)^{1/x} = [(1+ax)^{\frac{1}{ax}}]^a\) and then use the limit formula, (if you can use) \(\lim \limits_{x\to \infty} (1+1/x)^x = e\)
Brilliant! But the limit would then be 1 not 0...
Which according to wolfram it's 0.
http://www.wolframalpha.com/input/?i=lim+n-%3E+infinity+ln%281%2B%284-sin%28x%29%29%2Fn%29%5En
Ohh wait. I goofed. Have to take the ln of that.
ln(1) is 0. Thank you!
in the wolf, its shows ln^n
Yeah it's a notational thing. That just the inside raised to the n.
Kinda like sin^2(x) and (sin(x))^2.
okk... i thought the answer would be 4- sin x ..
Nah. It's 0.
:)
@hartnn : So I got up here.
\[\lim_{b \rightarrow 0}(1+(4-\sin(x)b)^{\frac{ 1 }{ b }}\]
Is that okay so far or am I way off?
i assume there is ln outside of that limit and you just plugged in b =1/n
Indeed sir.
yes, go on
Stuck >.< .
Like I know that should be e but I'm having trouble relating it to the definition.
whatever expression is with \(\Large 1+ ...\) that same expression should be with \(\Large \dfrac{1}{...}\) thats how I remember so we have 1+ (4-sin x)b so the fraction in the exponent should be \(\Large \dfrac{1}{(4-\sin x)b}\)
|dw:1443855809053:dw|
But the definition of e is (1+1/x)^x right? Here we have (1+(constant)b)^(1/b) . Are those equivalent?
http://www.wolframalpha.com/input/?i=lim+x-%3E+infty+%281%2Bax%29%5E%281%2F%28ax%29%29+
that will clear your doubt
Interesting. I did not know that even after 4 years of calculus and differential equations lol. I learn new things every day!
I want to learn how to solve this as well.. I get some steps but Im confused on others. :(
We take that as a formula, but its easy to prove that using L'Hopital's rule.
I will be writing out all the steps from the beginning
Yes I know we can use L'hopital's rule but for this assignment they (Other students) can't use that.
*drawing Let 4 - sin x = a , since its a constant. |dw:1443856327983:dw|
Yep I got that.
writing these steps for everyone's benefit :) |dw:1443856406335:dw|
Yep makes sense so far...
|dw:1443856506563:dw|
that big bracket evaluates to 1 1^a = 1 ln 1 = 0 :D
Cool!
Is there somewhere I can find the proof of the stuff in the brackets?
|dw:1443856722618:dw|
Yeah to get into the proper form for the limit.
lets prove it :) I'll use L'Hopitals, need to search the net for other proofs. |dw:1443856725213:dw| quick check, 0/0 form ln (1+ab) = ln 1 = 0 ab = 0 so we can apply L'Hopital's rule here
Nice!
we can do all kinds of mathematically legal manipulations to bring an expression in the standard form. i needed a form like (1+x)^(1/x) thats why I multiplied and divided by 'a' , which should be NON-ZERO (point to be noted.)
oh I see I see
How does this 0, the power would be undefined |dw:1443857378987:dw|
Oooh wait nvm, n = 1/b when n-> infinity, b ->9
b->0*
Yeah, there you go
Haha, I totally missed that, ok it's good now. Great explanation @hartnn thanks
me 4.
|dw:1443858029848:dw|
http://www.wolframalpha.com/input/?i=lim+n-%3E+infty+%281%2Ba%2Fn%29%5E%28%28n%29%29+ ln e^a = a ln e = a a = 4-sin x thats what I first got. but this wolf answer got me all confused and I ended up using b->infty instead of b->0
But that's wrong though. b goes to 0, not n.
not infinity*
can you explain as to `why` b \(\rightarrow\) 0 and not b \(\rightarrow \infty\) ?
true, and it makes sense logically too, constant/n = very very small no. 1+ very very small no. = 1 1^ very very larger number = 1 ln 1 =0 so the limit must go to 0 but with all the mathematical steps, I still get the answer as 4-sin x
It's 0 according to wolf :( .
Asking around. Like my feel is that the inside of that logarithm should be a 1.
Only then can we get a 0.
inside of a logarithm http://www.wolframalpha.com/input/?i=lim+n-%3E+infty+%281%2B%284-sin+x%29%2Fn%29%5E%28%28n%29%29+
It's witchcraft I tell you!
This is the statement of t he original problem.
1 Attachment
we can only bring limit inside a function if that function is continuous. and logarithm is indeed continuous...
http://www.wolframalpha.com/input/?i=lim+n-%3E+infinity+n*+ln+%281%2B%284-sin%28x%29%29%2Fn%29 even that gives 4-sin x!
haha oh my goodness x_x
Ohh wow. Wolfram is apparantly wrong.
http://math.stackexchange.com/questions/1462100/evaluating-the-limit-lim-n-rightarrow-infty-ln-left-1-frac-4-sinx
LOL
"which is apparently related to e" xD
http://i.imgur.com/F4xr6yp.png
Okay wow wolfram can't read notation clearly -.- .
its very rare case where wolfram goes wrong
Wow ._. ...
\(\color{blue}{\text{Originally Posted by}}\) @hartnn okk... i thought the answer would be 4- sin x .. \(\color{blue}{\text{End of Quote}}\) and then wasted an hour :P
All because we hail our god wolfram alpha too much ._. ...
xD It hardly fails!!! As humans we value consistency and dependability :P

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