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?
 A: Let's start with finding the limit of the reciprocal of the original expression:
\begin{align}
& \left(1 + \frac{1}{x}\right)^x \\
\end{align}
whose limit is the well-known $e$. Now you can conclude easily.

For those who needed more details, let me put it more explictly:
$$\left(\frac{x}{x + 1}\right)^x = \left(\frac{1}{1 + \frac{1}{x}}\right)^x = \frac{1}{\left(1 + \frac{1}{x}\right)^x}.$$
Now use $$\lim_{x \to \infty}\frac{h(x)}{g(x)} = \frac{\lim\limits_{x \to \infty}h(x)}{\lim\limits_{x \to \infty} g(x)}$$
when $\lim_{x \to \infty} g(x) \neq 0$. We conclude that
$$\lim_{x \to \infty}\left(\frac{x}{x + 1}\right)^x = \frac{1}{\lim\limits_{x \to \infty} \left(1 + \frac{1}{x}\right)^x} = \frac{1}{e}.$$
A: There are a variety of approaches to evaluate the limit of interest.  I thought that it would be instructive to show a way forward that uses only the squeeze theorem and standard inequalities.  To that end we proceed.
I showed in THIS ANSWER and THIS ONE that the logarithm function satisfies the inequalities for $z>0$
$$\frac{z-1}{z}\le \log z \le z-1 \tag 1$$

Another way to obtain $(1)$, is to rely on the integral definition of the logarithm expressed as 
$$\log z=\int_1^z \frac{1}{u}\,du$$

Now let $z=\frac{x}{x+1}$.  Then, we have
$$-1=x\,\left(\frac{\left(\frac{x}{x+1}-1\right)}{\frac{x}{x+1}}\right) \le x\,\log \left(\frac{x}{x+1}\right)\le x\,\left(\frac{x}{x+1}-1\right)=-\frac{x}{x+1}$$
Using the squeeze theorem, we see that 
$$\lim_{x\to \infty}x\,\log\left(\frac{x}{x+1}\right)=-1$$
Finally, using continuity of the exponential function reveals
$$\begin{align}
\lim_{x\to \infty}\left(\frac{x}{x+1}\right)^x&=\lim_{x\to \infty}e^{x\log \left(\frac{x}{x+1}\right)}\\\\
&=e^{\lim_{x\to \infty}x\log \left(\frac{x}{x+1}\right)}\\\\
&=e^{-1}
\end{align}$$
A: Another way to get it.
Consider $$A=\left(\frac{x}{x+1}\right)^x$$ and take logarithms $$\log(A)=x \log\left(\frac{x}{x+1}\right)=x \log\left(\frac{x+1-1}{x+1}\right)=x \log\left(1-\frac{1}{x+1}\right)$$ Now, remember that, for small $y$, $\log(1-y)\approx -y$. Replace $y$ by $\frac{1}{x+1}$ which makes $$\log(A)\approx -\frac{x}{x+1}=-\frac{1}{1+\frac 1x}\to -1$$ and then $A\to e^{-1}$
A: $\lim_{x\to\infty} (\frac{x}{(x+1)})^x\\
=\lim_{x\to\infty}(1-\frac{1}{(x+1)})^{(x+1)-1}\\
=\lim_{x\to\infty}(1-\frac{1}{(x+1)})^{(x+1)}\lim_{x\to\infty}(1-\frac{1}{(x+1)})^{-1}\\
=e^{-1}\cdot 1\\
=e^{-1}$
Notice you should transform your expression to well-known form.
A: 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.)
