$ \lim_{x \to \infty} (\frac {x+1}{x+2})^x $ For the limit
$ \lim_{x \to \infty} (\frac {x+1}{x+2})^x $
could you split it up into the fraction $ \lim_{x \to \infty} (1 - \frac{1}{x+2})^x$
and apply the standard limit $ \lim_{x\to+\infty} \left(1+\frac{k}{x}\right)^x=e^k $
or would you have to apply L'Hopitals rule
edit:
oops, forgot to add a term

$$\left( \frac{(x+2) - 1}{x+2} \right)^x = \left( 1 - \frac{1}{x+2} \right)^{x+2} \cdot \left( 1 - \frac{1}{x+2} \right)^{-2} \to e^{-1} \cdot 1 $$
