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

- anonymous

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

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- anonymous

\[\lim_{x \rightarrow 0} (1+1/2x)^x\]

- slaaibak

rewrite it as: e^(ln(1 + 1/2x)x)
then manipulate it into l'hopital

- 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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## More answers

- slaaibak

I'll quickly get a pen and try

- slaaibak

Just making sure, it's 2x in the denominator, right?

- anonymous

yeah is 1 over 2*x. this is why it is bugging me cuz i can't get rid of the indetermination

- anonymous

is it
\[\lim_{x\to \infty}(1+\frac{1}{2x})^x\]?

- anonymous

yeah

- slaaibak

you mean the limit to zero?

- anonymous

ok think of it as
\[(1+\frac{1}{2}\times \frac{1}{x})^x\]

- anonymous

as x tends to 0

- anonymous

oops ok lets try again

- anonymous

and no l'hopital??

- slaaibak

lol apparently not hey...

- TuringTest

that is cruel

- anonymous

ikr... such a wonderful tool. is like deriving using the concept of rate of change

- anonymous

ok maybe replace \(x\) by \(\frac{1}{z}\)

- anonymous

then take the limit as z goes to infinity

- anonymous

ok, so using l'hôspital, is the answer 1?

- slaaibak

I think I got it. Will explain now

- anonymous

oh lord i am in idiot!

- TuringTest

but then we get\[\exp\left(\lim_{z\to \infty}{\ln(1+\frac z2)\over z}\right)\]

- TuringTest

if you're an idiot satellite I hate to think what that makes me !

- anonymous

\[e^{x\ln(1+\frac{1}{2x}}=e^{x\ln(\frac{2x+1}{2x})}\]

- anonymous

tak the limit as x goes to zero oh wait nope not so snappy, ok i am not such an idiot

- slaaibak

That will only result in another indeterminate form

- anonymous

right i see that. for a moment i thought it was \(\ln(1)=0\) but it is not damn!

- slaaibak

kk I got it, will post method now:

- TuringTest

go slaiibak!

- anonymous

can't wait! no derivatives right?

- 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

lol that is what i got the first try, but i thought it was as x goes to infinity too !

- slaaibak

It does have a derivative in, but it's not L"Hospital. Or it's arguable, but will post quickly..

- anonymous

@satellite73 LOL

- anonymous

good lord... i'll have to spend these two days studying my retriceoff. have test on wednesday

- anonymous

whould it help to work the problem backwards? (from the answer)

- TuringTest

I doubt it

- 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

we can all get the answer with l'hospital, but you said \(no\) \(l'hospital\) so we are doing loops for you

- slaaibak

f'(x) = 2x(ln(2x+1) + x^2 / (2x + 1)
f'(0) = 0
therefore e^0 = 1.
So the limit = 1

- TuringTest

well, that's a proof all right
I guess since it doesn't technically use l'Hospital it's okay
nice one!

- 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

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

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

wow that was pretty snazzy!

- anonymous

haha, got my english from a brit school (I am brazillian)

- 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

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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