Question

(1-(3/x))^x x-> infinity solve using l'hoptials rule

Answer 1

@amistre64 I don't know how to solve it. Please, help

Answer 2

ln(1-3/x)
--------- = ln y
1/x

as x to infinity we get 0/0 right?

Answer 3

ln(1-3x^-1)
-----------
x^-1



3x^-2
---------------
-x^-2 (1-3x^-1)



-3
-------- = -3 = ln(y)
1- 3/x

but id have to check the wolf to be sure

Answer 4

http://www.wolframalpha.com/input/?i=limit%28x+to+inf%29+%281-3%2Fx%29%5Ex

yeah, e^-3

-3 = ln(y)

e^-3 = y

Answer 5

nope @amistre64 the problem is
\[lim_{x\rightarrow \infty} (1-\dfrac{3}{x})^x\]

Answer 6

thats what i did ....

Answer 7

a^x = y

x lna = lny

lna
---- = lny
1/x

etc

Answer 8

im confused how you got from the first step to the second one, can you explain it to me thank you

Answer 9

lhop works on indeterminate forms: 0/0

Answer 10

nvm , I got you

Answer 11

\[3x=\frac{3}{1/x}\]

Answer 12

i didnt label the steps, can you be more specfic?

Answer 13

openstudy is not playing well on my end .... gotta love a broken site

Answer 14

never mind i understand now thank you very much, i was confused because i didnt know you applied l'hoptials rule in your steps you were taking the derivative of hte bottom and top i was a bit confused. thank you very much!

Answer 15

\[\lim_{x \to a}f^x(x)\]
let f^x(x) = y, such that:
\[f^x(x)=y\]
\[x~ln(f(x))=ln(y)\]
\[\frac{ln(f(x))}{x^{-1}}=ln(y)\]now lHop if it takes the indeterminant form
\[\frac{f'(x)}{-x^{-2}~f(x)}\]
when you find the limit, then reverse the ln stuff

Answer 16

:) youre welcome