Question

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

Answer 1

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

Answer 2

any work to go with it?

Answer 3

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

Answer 4

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

Answer 5

sorry it is e^2

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

you other one is e^3

Answer 10

for exactly the same reason

Answer 11

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

Answer 12

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

Answer 13

sure,

Answer 14

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

Answer 15

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)}

Answer 16

oh wiat nvm

Answer 17

wait. myininaya i think it is just e^3

Answer 18

we do it the donkey way

Answer 19

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

Answer 20

which one do you want to do?

Answer 21

u r right

Answer 22

its not fair u shouldn't always win

Answer 23

fancy which one would you like worked out?

Answer 24

lim 3^-x x-> +(infinty

Answer 25

i want to see both :)

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

but we knock it out

Answer 30

why 0

Answer 31

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

Answer 32

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

Answer 33

limx-.2+ ln(x-2)

Answer 34

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

Answer 35

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

Answer 36

now we have 0/0 so use l'hopital

Answer 37

ooo ojkak

Answer 38

ooo ojkak

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

then let x -> infinity and get 2

Answer 43

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

Answer 44

so many questions, so little time