Question

Solve: lim x->infinity (x/(1+x))^x

Answer 1

To Clarify: \[\lim_{x \rightarrow \infty}\left( \frac{ x }{ x +1} \right)^{x} \]

Answer 2

which could go to \[\lim_{x \rightarrow \infty}\epsilon^{xlnx - \ln(x+1) }\]

Answer 3

\[\lim_{x\to\infty}\left(\frac{x}{x+1}\right)^x=\lim_{x\to\infty}\exp\left[\ln\left[\left(\frac{x}{x+1}\right)^x\right]\right]\]
Using the continuity of the exponential function, you have
\[\exp\left[\lim_{x\to\infty}x\ln\left(\frac{x}{x+1}\right)\right]\]
The limit above gives you the indeterminate form \(\infty\cdot0\), so rewrite a bit to get
\[\exp\left[\lim_{x\to\infty}\frac{\ln\left(\frac{x}{x+1}\right)}{\frac{1}{x}}\right]\]
Now, the limit gives you indeterminate form \(\dfrac{0}{0}\). Apply L'Hopital's rule as many times as necessary.

Answer 4

Ok, I know the answer is \[e ^{-1}\].. Will give it a shot. Thanks man.

Answer 5

You're welcome!

Answer 6

L'hopitals, take the limit of the same function. Just denominator and numerator derived yes?

Answer 7

Yes.

Answer 8

\[\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}\]
given that the left limit is 0/0 or ±∞/∞.

Answer 9

so this would give me exp \[\exp\left[ \lim_{x \rightarrow \infty} \frac{ x }{ 1+x }\right]\]

Answer 10

Which would be e^0 which is wrong..

Answer 11

it would be \(e^1\) which would still be wrong. ;)

Answer 12

My second mistake then, Yay :P

Answer 13

Numerator:\[\frac{d}{dx}\ln\left(\frac{x}{x+1}\right)=\frac{\frac{x+1-x}{(x+1)^2}}{\frac{x}{x+1}}=\frac{1}{x(x+1)}\]
Denominator:
\[\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}\]
\[\frac{\frac{1}{x(x+1)}}{-\frac{1}{x^2}}=\color{red}-\frac{x}{x+1}\]

Answer 14

So thats where the neg is hiding, thank you :)

Answer 15

One last question, \[\lim_{x \rightarrow \infty } -\frac{ x }{ x + 1}\] is -1 because the +1 in the denominator becomes negligible yes?

Answer 16

You could think of it that way, yeah. If you divide by each term in numerator and denominator by \(x\), you have
\[-\frac{1}{1+\frac{1}{x}}\]
The \(\dfrac{1}{x}\) in the denominator approaches 0 as x approaches infinity.