anonymous
  • anonymous
want some help with this Limit lim(x->0) (1+1/(2*x))^x at least a guidance! thanks!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\lim_{x \rightarrow 0} (1+1/2x)^x\]
slaaibak
  • slaaibak
rewrite it as: e^(ln(1 + 1/2x)x) then manipulate it into l'hopital
anonymous
  • anonymous
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?

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slaaibak
  • slaaibak
I'll quickly get a pen and try
slaaibak
  • slaaibak
Just making sure, it's 2x in the denominator, right?
anonymous
  • anonymous
yeah is 1 over 2*x. this is why it is bugging me cuz i can't get rid of the indetermination
anonymous
  • anonymous
is it \[\lim_{x\to \infty}(1+\frac{1}{2x})^x\]?
anonymous
  • anonymous
yeah
slaaibak
  • slaaibak
you mean the limit to zero?
anonymous
  • anonymous
ok think of it as \[(1+\frac{1}{2}\times \frac{1}{x})^x\]
anonymous
  • anonymous
as x tends to 0
anonymous
  • anonymous
oops ok lets try again
anonymous
  • anonymous
and no l'hopital??
slaaibak
  • slaaibak
lol apparently not hey...
TuringTest
  • TuringTest
that is cruel
anonymous
  • anonymous
ikr... such a wonderful tool. is like deriving using the concept of rate of change
anonymous
  • anonymous
ok maybe replace \(x\) by \(\frac{1}{z}\)
anonymous
  • anonymous
then take the limit as z goes to infinity
anonymous
  • anonymous
ok, so using l'hôspital, is the answer 1?
slaaibak
  • slaaibak
I think I got it. Will explain now
anonymous
  • anonymous
oh lord i am in idiot!
TuringTest
  • TuringTest
but then we get\[\exp\left(\lim_{z\to \infty}{\ln(1+\frac z2)\over z}\right)\]
TuringTest
  • TuringTest
if you're an idiot satellite I hate to think what that makes me !
anonymous
  • anonymous
\[e^{x\ln(1+\frac{1}{2x}}=e^{x\ln(\frac{2x+1}{2x})}\]
anonymous
  • anonymous
tak the limit as x goes to zero oh wait nope not so snappy, ok i am not such an idiot
slaaibak
  • slaaibak
That will only result in another indeterminate form
anonymous
  • anonymous
right i see that. for a moment i thought it was \(\ln(1)=0\) but it is not damn!
slaaibak
  • slaaibak
kk I got it, will post method now:
TuringTest
  • TuringTest
go slaiibak!
anonymous
  • anonymous
can't wait! no derivatives right?
anonymous
  • anonymous
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....
anonymous
  • anonymous
lol that is what i got the first try, but i thought it was as x goes to infinity too !
slaaibak
  • slaaibak
It does have a derivative in, but it's not L"Hospital. Or it's arguable, but will post quickly..
anonymous
  • anonymous
@satellite73 LOL
anonymous
  • anonymous
good lord... i'll have to spend these two days studying my retriceoff. have test on wednesday
anonymous
  • anonymous
whould it help to work the problem backwards? (from the answer)
TuringTest
  • TuringTest
I doubt it
slaaibak
  • slaaibak
\[\exp(xln(2x+1)) = \exp({x^2 \ln(2x + 1) \over x})\] 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
  • TuringTest
we can all get the answer with l'hospital, but you said \(no\) \(l'hospital\) so we are doing loops for you
slaaibak
  • slaaibak
f'(x) = 2x(ln(2x+1) + x^2 / (2x + 1) f'(0) = 0 therefore e^0 = 1. So the limit = 1
TuringTest
  • TuringTest
well, that's a proof all right I guess since it doesn't technically use l'Hospital it's okay nice one!
anonymous
  • anonymous
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
slaaibak
  • slaaibak
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.
anonymous
  • anonymous
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!!
anonymous
  • anonymous
wow that was pretty snazzy!
anonymous
  • anonymous
haha, got my english from a brit school (I am brazillian)
slaaibak
  • slaaibak
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!
anonymous
  • anonymous
thanks! 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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