Question

Limits,

Can you help me solve this limit: http://screencast.com/t/ChQdqMNFKB

Answer 1

\[\lim_{x\to\infty}\left(1-\frac{3}{x}\right)^x=e^{\displaystyle\ln\lim_{x\to\infty}\left(1-\frac{3}{x}\right)^x}\]
Since \(\ln x\) is continuous as \(x\to\infty\), you have
\[e^{\displaystyle\lim_{x\to\infty}\ln\left(1-\frac{3}{x}\right)^x}\]
Then apply some log properties:
\[e^{\displaystyle\lim_{x\to\infty}x\ln\left(1-\frac{3}{x}\right)}=e^{\displaystyle\lim_{x\to\infty}\frac{\ln\left(1-\frac{3}{x}\right)}{\frac{1}{x}}}\]
A candidate for l'Hopital's rule.

Answer 2

another way,
let \(\large \frac{-3}{x} = y\)

Answer 3

All of those are the appropriate methods, I just thought Id throw something random in. This limitis pretty close to what you could kinda say is an identity, just not exactly. A form of the compound interest formula:

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

Answer 4

@Christos use the latter if you don't know L'Hopital's rule.

Answer 5

So the limit is,
\[e^{-3}\]

Answer 6

@Christos do you solve it right?