anyone good at limits?
lim(1+2/x)^x
x-(infinity)

- anonymous

anyone good at limits?
lim(1+2/x)^x
x-(infinity)

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- anonymous

\[e^{-2}\]is my guess

- anonymous

any work to go with it?

- myininaya

he might be right i remember that limit i think it was either as x->0 or infinity cant remember

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## More answers

- anonymous

lim(1+3h)^(1/h)
h->0

- anonymous

sorry it is e^2

- anonymous

\[e=lim_{x->\infty}(1+\frac{1}{x})^x\]

- myininaya

if he is right about that one then this second one should be e^(1/[3h]) i believe

- anonymous

and so of course
\[e^2=lim_{x->\infty}(1+\frac{2}{x})^x\]

- anonymous

you other one is e^3

- anonymous

for exactly the same reason

- anonymous

so then lim 3^(-x)
x->+(infinity)

- anonymous

if you want we can work this out step by step.

- anonymous

sure,

- anonymous

step one is take the log get
\[ln(1+\frac{1}{x})^x=xln(1+\frac{1}{x})\]

- myininaya

let 1/u=3h
as h->0 then u->infinity
so we have as u->infinity then (1+1/u)^u->e^u but u=1/(3h)
so the limit is e^{1/(3h)}

- myininaya

oh wiat nvm

- anonymous

wait. myininaya i think it is just e^3

- anonymous

we do it the donkey way

- anonymous

take the log, take the limit using l'hopital, see that we get 3 and then conclude that it is e^3

- anonymous

which one do you want to do?

- myininaya

u r right

- myininaya

its not fair u shouldn't always win

- anonymous

fancy which one would you like worked out?

- anonymous

lim 3^-x x-> +(infinty

- myininaya

i want to see both :)

- anonymous

\[lim_{x->\infty}3^{-x}\]

- anonymous

nothing to that one. that is
\[lim_{x->\infty}\frac{1}{3^x}=0\]

- anonymous

this is the pain:
(1+2/x)^x

- anonymous

but we knock it out

- anonymous

why 0

- anonymous

as x gets large 3^x gets really large. huge denominator,1 in the numerator, very close to zero

- anonymous

back to (1+2/x)^x because i gotta go

- anonymous

limx-.2+ ln(x-2)

- anonymous

take the log get xln(1+2/x)
take the limit as x -> infinity get infinity times 0 so rewrite

- anonymous

as \[\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}\]

- anonymous

now we have 0/0 so use l'hopital

- anonymous

ooo ojkak

- anonymous

ooo ojkak

- anonymous

i made a mistake and wrote 1 instead of 2. when you take the derivative of to bottom you get
\[-\frac{1}{x^2}\]

- anonymous

when you take the derivative of the top you get
\[\frac{1}{1+\frac{2}{x}}\times-2\frac{1}{x^2}\]

- anonymous

everything cancels leaving you with \[\frac{2}{1+\frac{1}{x^2}}\]

- anonymous

then let x -> infinity and get 2

- anonymous

and since we took the log in the first step the answer is not 2, but e^2

- anonymous

so many questions, so little time

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