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

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

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

GChamonBest ResponseYou've already chosen the best response.0
thing 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?
 2 years ago

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

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

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

satellite73Best ResponseYou've already chosen the best response.0
is it \[\lim_{x\to \infty}(1+\frac{1}{2x})^x\]?
 2 years ago

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

satellite73Best ResponseYou've already chosen the best response.0
ok think of it as \[(1+\frac{1}{2}\times \frac{1}{x})^x\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
oops ok lets try again
 2 years ago

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

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

satellite73Best ResponseYou've already chosen the best response.0
ok maybe replace \(x\) by \(\frac{1}{z}\)
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
then take the limit as z goes to infinity
 2 years ago

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

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

satellite73Best ResponseYou've already chosen the best response.0
oh lord i am in idiot!
 2 years ago

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

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

satellite73Best ResponseYou've already chosen the best response.0
\[e^{x\ln(1+\frac{1}{2x}}=e^{x\ln(\frac{2x+1}{2x})}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
tak the limit as x goes to zero oh wait nope not so snappy, ok i am not such an idiot
 2 years ago

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

satellite73Best ResponseYou've already chosen the best response.0
right i see that. for a moment i thought it was \(\ln(1)=0\) but it is not damn!
 2 years ago

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

satellite73Best ResponseYou've already chosen the best response.0
can't wait! no derivatives right?
 2 years ago

KreshnikBest ResponseYou've already chosen the best response.0
You'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....
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
lol that is what i got the first try, but i thought it was as x goes to infinity too !
 2 years ago

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

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

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

slaaibakBest 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 \[\exp( \lim_{x \rightarrow 0} {x^2\ln(2x + 1)  (0)^2 \ln(2(0) + 1)\over x  0} ) = \exp(f'(0))\]
 2 years ago

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

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

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

GChamonBest ResponseYou've already chosen the best response.0
would 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
 2 years ago

slaaibakBest ResponseYou've already chosen the best response.4
It'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.
 2 years ago

GChamonBest ResponseYou've already chosen the best response.0
alright! 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!!
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
wow that was pretty snazzy!
 2 years ago

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

slaaibakBest ResponseYou've already chosen the best response.4
GChamon, 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!
 2 years ago

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