At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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

any work to go with it?

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

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

sorry it is e^2

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

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

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

you other one is e^3

for exactly the same reason

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

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

sure,

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

oh wiat nvm

wait. myininaya i think it is just e^3

we do it the donkey way

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

which one do you want to do?

u r right

its not fair u shouldn't always win

fancy which one would you like worked out?

lim 3^-x x-> +(infinty

i want to see both :)

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

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

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

but we knock it out

why 0

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

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

limx-.2+ ln(x-2)

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

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

now we have 0/0 so use l'hopital

ooo ojkak

ooo ojkak

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

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

then let x -> infinity and get 2

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

so many questions, so little time