Find the limit $\lim\limits_{x\to\infty} f(x)= \lim\limits_{x \to \infty} \left(\frac{x}{x+1}\right)^x$ I need to find the following:
$$\lim_{x\to\infty} f(x)= \lim_{x \to \infty}\left (\frac{x}{x+1} \right )^x$$
I know that this limit = $\frac{1}{e}$ from plugging it into a calculator, but I have to prove it without using the fact that:
$$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}\left (\frac{x}{x+k} \right )^x=\frac{1}{e^k}$$
I started by exponentiating:
$$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{\ln \left (\frac{x}{x+1} \right )^x}$$
and from here I've dropped the exponent $x$ in front of the $\ln$, and from here I'm getting stuck. Should I separate the $\ln$ like this?
$$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{x(\ln(x)-\ln(x+1))}$$
This doesn't seem to be leading me down the right path, but I'm not sure how else to do it. Is there a way to apply L'Hopital? If so, how?

First you set the entire equation equal to y:
$$
(\frac {x}{(x+1)})^x = y
$$
We can then insert both sides into $ln(x)$:
$$
ln((\frac x{(x+1)})^x) = ln(y)
$$
Then pull out the x power:
$$
xln(\frac x{(x+1)}) = ln(y)
$$
Split the natural log:
$$
xln(x) - xln(x+1) = ln(y)
$$
Shift the left term into the bottom of a fraction and add it to the other term:
$$
\frac {1}{\frac 1{xln(x)}} - xln(x+1) = ln(y)
$$
$$
\frac {1}{\frac 1{xln(x)}} - \frac {\frac {xln(x+1)}{xln(x)}}{\frac 1{xln(x)}} = ln(y)
$$
$$
\frac {1 - \frac {ln(x+1)}{ln(x)}}{\frac 1{xln(x)}} = ln(y)
$$
Taking the limits of the top and bottom yields the form $\frac 00$. Therefore, L'hopetal's Rule can now be applied:
$$
\frac {\frac {(x+1)ln(x)-xln(x+1)}{ln^2(x)}}{\frac {ln(x) + \frac xx}{(xln(x))^2}} = ln(y)
$$
A little simplification and canceling...
$$
\frac {x^3ln(x)+x^2ln(x)-x^3ln(x+1)}{ln(x) + 1} = ln(y)
$$
L'Hopetals rule still applies.
$$
\frac {2x^2ln(x) + \frac {x^3}x + 2xln(x) + \frac {x^2}x - 3x^2ln(x+1) - \frac {x^3}{x+1}}{\frac 1x} = ln(y)
$$
Simplify:
$$
2x^3ln(x) + x^3 + 2x^2ln(x) + x^2 - 3x^3ln(x+1) - \frac {x^4}{x+1} = ln(y)
$$
Put everything back under $x+1$
$$
\frac {(2x^3ln(x) + x^3 + 2x^2ln(x) + x^2 - 3x^3ln(x+1))(x+1)}{x+1} - \frac {x^4}{x+1} = ln(y)
$$
$$
\frac {2x^4ln(x) + 2x^3ln(x) + x^4 + x^3 + 2x^2ln(x) + 2x^3ln(x) + x^2 + x^3 - 3x^3ln(x+1) - 3x^4ln(x+1) - x^4}{x+1} = ln(y)
$$
$$
\frac {2x^4ln(x) + 4x^3ln(x) + 2x^3 + 2x^2ln(x) + x^2 - 3x^3ln(x+1) - 3x^4ln(x+1)}{x+1} = ln(y)
$$
$$
\frac {ln(x)(2x^4 + 4x^3 + 2x^2) + 2x^3 + x^2 - ln((x+1)(3x^3 + 3x^4)}{x+1} = ln(y)
$$
$$
\frac {ln((x)^{2x^4 + 4x^3 + 2x^2}) + 2x^3 + x^2 - ln((x+1)^{3x^3 + 3x^4}}{x+1} = ln(y)
$$
$$
\frac {ln(\frac{(x)^{2x^4 + 4x^3 + 2x^2}}{(x+1)^{3x^3 + 3x^4}}) + 2x^3 + x^2 }{x+1} = ln(y)
$$
$$
2x^2 + \frac {ln(\frac{(x)^{2x^4 + 4x^3 + 2x^2}}{(x+1)^{3x^3 + 3x^4}}) - x^2 }{x+1} = ln(y)
$$
This will continue on and on until you get $\frac 1e$
At this point I am stuck, but I believe I have well illustrated the way you would do this WITHOUT substituting values in other known limits. Unfortunately it is too late at night for me and I just can't see how to reduce it so that L'hopetals rule isn't infinite. (At least, it appears like it'd be applied infinitely from my perspective.)
