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GChamon
want some help with this Limit lim(x->0) (1+1/(2*x))^x at least a guidance! thanks!
\[\lim_{x \rightarrow 0} (1+1/2x)^x\]
rewrite it as: e^(ln(1 + 1/2x)x) then manipulate it into l'hopital
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?
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 \[\lim_{x\to \infty}(1+\frac{1}{2x})^x\]?
you mean the limit to zero?
ok think of it as \[(1+\frac{1}{2}\times \frac{1}{x})^x\]
oops ok lets try again
lol apparently not hey...
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:
can't wait! no derivatives right?
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....
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)
\[\exp(xln(2x+1)) = \exp({x^2 \ln(2x + 1) \over x})\] 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 \[\exp( \lim_{x \rightarrow 0} {x^2\ln(2x + 1) - (0)^2 \ln(2(0) + 1)\over x - 0} ) = \exp(f'(0))\]
we can all get the answer with l'hospital, but you said \(no\) \(l'hospital\) so we are doing loops for you
f'(x) = 2x(ln(2x+1) + x^2 / (2x + 1) f'(0) = 0 therefore e^0 = 1. So the limit = 1
well, that's a proof all right I guess since it doesn't technically use l'Hospital it's okay nice one!
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
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.
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!!
wow that was pretty snazzy!
haha, got my english from a brit school (I am brazillian)
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!
thanks! that cheers me little to study! they all told me engineering is not easy. starting to see they were all right