Question

limit. (exponent and lhospital)
limit (1-1/x)^(0.7x) as x->infinity
(formatting in reply)

Answer 1

eyeball it

Answer 2

\[\lim_{x \rightarrow \infty}(1-\frac{ 1 }{ x })^{0.7x}\]
Do I take a natural log? I'm not sure how to work this.

Answer 3

oh, ok, then (1 - 0)^0 ?

Answer 4

power rule

Answer 5

let me say it correctly
\[\lim_{x\to \infty}(1+\frac{1}{x})^x=e\] that is pretty much by definition
it follows that
\[\lim_{x\to \infty}(1-\frac{1}{x})^x=e^{-1}=\frac{1}{e}\]

Answer 6

and that
\[\lim_{x\to \infty}(1-\frac{1}{x})^{-7x}=e^{-.7}=\frac{1}{e^{.7}}\]

Answer 7

you can of course take the log, and use l'hopital for that, then exponentiate
that will probably make your math teacher happier

Answer 8

take the log, get
\[.7x\ln(1-\frac{1}{x})\] now you are in the form \(\infty\times 0\) usual trick is to rewrite as
\[\frac{\ln(1-\frac{1}{x})}{\frac{1}{.7x}}\]
then take the derivative top and bottom
after a bunch of boring algebra you get \(-.7\) so your answer will be \(e^{-.7}\)

Answer 9

\[\lim_{x \rightarrow \infty}(1+\frac{ 1 }{ x })^x = e\]
is a known rule that I should remember?

Answer 10

ok, that last part with the ln and LH looks more familiar, I'm going to work that out for a min.

Answer 11

what is your definition of \(e\)? sometimes that is used to define the number

Answer 12

yes, you can take the derivative top and bottom. it will be some annoying algebra
maybe easier if you write
\[\ln(1-\frac{1}{x})\] as
\[\ln(\frac{1-x}{x})\]

Answer 13

just that e = 2.72 something. wow, this will be a bunch of harsh algebra. How does e get introduced this way? (the ln / algebra route)