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\[\lim_{n \rightarrow \infty}\ln\left( 1+ \frac{ 4-\sin(x) }{ n } \right)^n\]

4 -sin x or 4 - sin n ??
if its 4- sin x, then its a constant...

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.

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.

:)

\[\lim_{b \rightarrow 0}(1+(4-\sin(x)b)^{\frac{ 1 }{ b }}\]

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.

|dw:1443855809053:dw|

http://www.wolframalpha.com/input/?i=lim+x-%3E+infty+%281%2Bax%29%5E%281%2F%28ax%29%29+

that will clear your doubt

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.

Nice!

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|

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 \(\rightarrow \infty\) ?

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.

It's witchcraft I tell you!

This is the statement of t he original problem.

haha oh my goodness x_x

Ohh wow. Wolfram is apparantly wrong.

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 ._. ...

All because we hail our god wolfram alpha too much ._. ...

xD It hardly fails!!! As humans we value consistency and dependability :P