What is the limit of $\left(\frac{\sqrt[n]{a}+b}{b+1}\right)^n$? Assume $a$ and $b$ are positive real numbers prove that:
$\lim_{n\rightarrow \infty} \left(\frac{\sqrt[n]{a}+b}{b+1}\right)^n = a^{1/(b+1)}$
Actually the main question was longer than this but the other states of the problem were easier but this part I am stuck...
any help would be appreciated.

Always remember that $x^y=\exp(y\log x)$. So we need to show that
$$\lim_{n\rightarrow \infty} \exp\left(n\cdot\log\frac{\sqrt[n]{a}+b}{b+1}\right) = \exp\left(\frac{\log a}{b+1}\right)\text.$$
Since $\exp$ is continuous, this is equivalent to
$$\lim_{n\rightarrow \infty} n\cdot\log\frac{\sqrt[n]{a}+b}{b+1} = \frac{\log a}{b+1}\text.$$
Proof:
Substituting $n=1/t$
$$\lim_{n\rightarrow \infty} n\cdot\log\frac{\sqrt[n]{a}+b}{b+1} = \lim_{t\rightarrow 0^+} \frac{\log\frac{\exp(t\cdot\log a)+b}{b+1}}t\underset{\text{l'Hôpital}}{=}\lim_{t\rightarrow 0^+} \frac{\left(\log\frac{\exp(t\cdot\log a)+b}{b+1}\right)'}{t'}=\\ 
\lim_{t\rightarrow 0^+} \left(\log\frac{\exp(t\cdot\log a)+b}{b+1}\right)'=\lim_{t\rightarrow 0^+} \frac{b+1}{\exp(t\cdot\log a)+b}\cdot\frac{\log a\cdot\exp(t\cdot\log a)}{b+1}=\frac{\log a}{b+1}$$
where the last equality follows by simply plugging in $t=0$, since $\exp$ is continuous in $0$.
