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

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

I'll quickly get a pen and try

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

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

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

yeah

you mean the limit to zero?

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

as x tends to 0

oops ok lets try again

and no l'hopital??

lol apparently not hey...

that is cruel

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

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

then take the limit as z goes to infinity

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

I think I got it. Will explain now

oh lord i am in idiot!

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

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

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

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

That will only result in another indeterminate form

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

kk I got it, will post method now:

go slaiibak!

can't wait! no derivatives right?

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

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

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

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

I doubt it

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

wow that was pretty snazzy!

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