How to solve $$\lim_{x \to \infty}(\dfrac{x}{x+1})^x$$
The answer is $\dfrac{1}{e}$
I can factor the $x$ out to get:
$$\lim_{x \to \infty}\left(\dfrac{x(1)}{x(1+1/x)}\right)^x = \lim_{x \to \infty}\left(\dfrac{1}{1+1/x)}\right)^x$$
How do I further simplify this to get to my limit?
On
This solution does not require L'Hopital. However, it does require the following daunting identity. To learn more about it (trust me, once you've used it once, you'll never want to forget it), see this.
$$\lim\limits_{x \to a}{\phi(x)^{\psi(x)}} = e^{\lim\limits_{x \to a}{[(\phi(x)-1)\psi(x)}]}$$
Using this identity...
$$\lim_{x \to \infty}\left(\dfrac{x}{x+1}\right)^x = e^{\lim_{x \to \infty}\left(\dfrac{x}{x+1} - 1\right)x} = e^{\lim_{x \to \infty}-\dfrac{x}{x+1}} = e^{-1} = \frac{1}{e}$$
You almost got it:
$$\left(\frac1{1+\frac1x}\right)^x=\frac1{\left(1+\frac1x\right)^x}\xrightarrow[x\to\infty]{}\frac1e$$
where the limit is gotten using arithmetic of limits...