Question

want some help with this Limit

lim(x->0) (1+1/(2*x))^x

at least a guidance! thanks!

Answer 1

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

Answer 2

rewrite it as: e^(ln(1 + 1/2x)x)

then manipulate it into l'hopital

Answer 3

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?

Answer 4

I'll quickly get a pen and try

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yeah

Answer 9

you mean the limit to zero?

Answer 10

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

Answer 11

as x tends to 0

Answer 12

oops ok lets try again

Answer 13

and no l'hopital??

Answer 14

lol apparently not hey...

Answer 15

that is cruel

Answer 16

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

Answer 17

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

Answer 18

then take the limit as z goes to infinity

Answer 19

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

Answer 20

I think I got it. Will explain now

Answer 21

oh lord i am in idiot!

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

That will only result in another indeterminate form

Answer 27

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

Answer 28

kk I got it, will post method now:

Answer 29

go slaiibak!

Answer 30

can't wait! no derivatives right?

Answer 31

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....

Answer 32

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

Answer 33

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

Answer 34

@satellite73 LOL

Answer 35

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

Answer 36

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

Answer 37

I doubt it

Answer 38

\[\exp(xln(2x+1)) = \exp({x^2 \ln(2x + 1) \over x})\]

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

Answer 39

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

Answer 40

f'(x) = 2x(ln(2x+1) + x^2 / (2x + 1)

f'(0) = 0

therefore e^0 = 1.

So the limit = 1

Answer 41

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

Answer 42

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

Answer 43

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.

Answer 44

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!!

Answer 45

wow that was pretty snazzy!

Answer 46

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

Answer 47

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!

Answer 48

thanks! that cheers me little to study!
they all told me engineering is not easy. starting to see they were all right