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anonymous
 4 years ago
want some help with this Limit
lim(x>0) (1+1/(2*x))^x
at least a guidance! thanks!
anonymous
 4 years ago
want some help with this Limit lim(x>0) (1+1/(2*x))^x at least a guidance! thanks!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0} (1+1/2x)^x\]

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4rewrite it as: e^(ln(1 + 1/2x)x) then manipulate it into l'hopital

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thing is, we still haven't got to l'hôspital even though I know what ur talkign about. Teacher is not gonna consider any answer using l'hôspital... a real setback to these exercises. anyway to solve this without this tool?

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4I'll quickly get a pen and try

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4Just making sure, it's 2x in the denominator, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah is 1 over 2*x. this is why it is bugging me cuz i can't get rid of the indetermination

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is it \[\lim_{x\to \infty}(1+\frac{1}{2x})^x\]?

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4you mean the limit to zero?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok think of it as \[(1+\frac{1}{2}\times \frac{1}{x})^x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oops ok lets try again

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4lol apparently not hey...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ikr... such a wonderful tool. is like deriving using the concept of rate of change

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok maybe replace \(x\) by \(\frac{1}{z}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then take the limit as z goes to infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, so using l'hôspital, is the answer 1?

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4I think I got it. Will explain now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh lord i am in idiot!

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0but then we get\[\exp\left(\lim_{z\to \infty}{\ln(1+\frac z2)\over z}\right)\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0if you're an idiot satellite I hate to think what that makes me !

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[e^{x\ln(1+\frac{1}{2x}}=e^{x\ln(\frac{2x+1}{2x})}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0tak the limit as x goes to zero oh wait nope not so snappy, ok i am not such an idiot

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4That will only result in another indeterminate form

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right i see that. for a moment i thought it was \(\ln(1)=0\) but it is not damn!

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4kk I got it, will post method now:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can't wait! no derivatives right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You're supposed to know this rule: \[\LARGE \lim_{x\to \infty}\left(1+\frac{1}{x}\right)^x=e\] \[\Large \lim_{x\to 0} \left(1+\frac{1}{2x}\right)^x\] \[\Large \lim_{x\to 0} \left[\left(1+\frac{1}{2x}\right)^{\frac{2x}{1}\cdot \frac{1}{2x }}\right]^x\] \[\Large \lim_{x\to 0} \left[\left(1+\frac{1}{2x}\right)^{2x} \right]^{\frac{x}{2x}}\] \[\LARGE \left(e\right)^{\frac12}=\sqrt e\] AAhhh.. LOOK WHAT I'VE DONE !! NOW I SEE THAT IT'S NOT INFINITE BUT IS "0" AUFFF... ( at least I'll post it :P ) hahahaah....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol that is what i got the first try, but i thought it was as x goes to infinity too !

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4It does have a derivative in, but it's not L"Hospital. Or it's arguable, but will post quickly..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0good lord... i'll have to spend these two days studying my retriceoff. have test on wednesday

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whould it help to work the problem backwards? (from the answer)

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4\[\exp(xln(2x+1)) = \exp({x^2 \ln(2x + 1) \over x})\] now using this: \[\lim_{x \rightarrow a} {f(x)  f(a) \over x  a} = f'(a)\] setting f(x) = x^2ln(2x + 1) we get \[\exp( \lim_{x \rightarrow 0} {x^2\ln(2x + 1)  (0)^2 \ln(2(0) + 1)\over x  0} ) = \exp(f'(0))\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0we can all get the answer with l'hospital, but you said \(no\) \(l'hospital\) so we are doing loops for you

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4f'(x) = 2x(ln(2x+1) + x^2 / (2x + 1) f'(0) = 0 therefore e^0 = 1. So the limit = 1

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0well, that's a proof all right I guess since it doesn't technically use l'Hospital it's okay nice one!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would anyone bother to talk me a little bit through the proof? exp(xln(2x+1))=exp(x2ln(2x+1)x) how do u get to this in the first place? i know it sounds dumb but i am new to calculus

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4It's just a little bit of algebraic manipulation to force it into the f'(a) form. Check if you simplify, you will get back to the original thing.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright! gonna put some effort into learning more about functions since it is somehow not so clear to me how to manipulate the original expression. nevertheless, surely u guys helped a lot! thanks!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow that was pretty snazzy!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha, got my english from a brit school (I am brazillian)

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.4GChamon, as you do more problems and exercises, you'll start to see it much clearer. Since you're only in the start of calculus now, you'll learn so much more and these things will become more intuitive. Good luck with the test!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks! that cheers me little to study! they all told me engineering is not easy. starting to see they were all right
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