$ \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

Hint
Let $$A= \Big(\frac {x+1}{x+2}\Big)^x$$ and take logarithm; so $$\log(A)=x\log\Big(\frac {x+1}{x+2}\Big)$$ Perform the long division and get $$\frac {x+1}{x+2}=1-\frac{1}{x}+\frac{2}{x^2}+\cdots$$ Now, use the fact that, for small $y$, $\log(1+y)\approx y$
I am sure that you can take from here.
