How to solve $\lim\limits_{x \to \infty}\left(\frac{x}{x+1}\right)^x$ 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?

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...
