anonymous
  • anonymous
anyone good at limits? lim(1+2/x)^x x-(infinity)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[e^{-2}\]is my guess
anonymous
  • anonymous
any work to go with it?
myininaya
  • 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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anonymous
  • anonymous
lim(1+3h)^(1/h) h->0
anonymous
  • anonymous
sorry it is e^2
anonymous
  • anonymous
\[e=lim_{x->\infty}(1+\frac{1}{x})^x\]
myininaya
  • myininaya
if he is right about that one then this second one should be e^(1/[3h]) i believe
anonymous
  • anonymous
and so of course \[e^2=lim_{x->\infty}(1+\frac{2}{x})^x\]
anonymous
  • anonymous
you other one is e^3
anonymous
  • anonymous
for exactly the same reason
anonymous
  • anonymous
so then lim 3^(-x) x->+(infinity)
anonymous
  • anonymous
if you want we can work this out step by step.
anonymous
  • anonymous
sure,
anonymous
  • anonymous
step one is take the log get \[ln(1+\frac{1}{x})^x=xln(1+\frac{1}{x})\]
myininaya
  • 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
  • myininaya
oh wiat nvm
anonymous
  • anonymous
wait. myininaya i think it is just e^3
anonymous
  • anonymous
we do it the donkey way
anonymous
  • anonymous
take the log, take the limit using l'hopital, see that we get 3 and then conclude that it is e^3
anonymous
  • anonymous
which one do you want to do?
myininaya
  • myininaya
u r right
myininaya
  • myininaya
its not fair u shouldn't always win
anonymous
  • anonymous
fancy which one would you like worked out?
anonymous
  • anonymous
lim 3^-x x-> +(infinty
myininaya
  • myininaya
i want to see both :)
anonymous
  • anonymous
\[lim_{x->\infty}3^{-x}\]
anonymous
  • anonymous
nothing to that one. that is \[lim_{x->\infty}\frac{1}{3^x}=0\]
anonymous
  • anonymous
this is the pain: (1+2/x)^x
anonymous
  • anonymous
but we knock it out
anonymous
  • anonymous
why 0
anonymous
  • anonymous
as x gets large 3^x gets really large. huge denominator,1 in the numerator, very close to zero
anonymous
  • anonymous
back to (1+2/x)^x because i gotta go
anonymous
  • anonymous
limx-.2+ ln(x-2)
anonymous
  • anonymous
take the log get xln(1+2/x) take the limit as x -> infinity get infinity times 0 so rewrite
anonymous
  • anonymous
as \[\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}\]
anonymous
  • anonymous
now we have 0/0 so use l'hopital
anonymous
  • anonymous
ooo ojkak
anonymous
  • anonymous
ooo ojkak
anonymous
  • 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
  • anonymous
when you take the derivative of the top you get \[\frac{1}{1+\frac{2}{x}}\times-2\frac{1}{x^2}\]
anonymous
  • anonymous
everything cancels leaving you with \[\frac{2}{1+\frac{1}{x^2}}\]
anonymous
  • anonymous
then let x -> infinity and get 2
anonymous
  • anonymous
and since we took the log in the first step the answer is not 2, but e^2
anonymous
  • anonymous
so many questions, so little time

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