$ \lim_{n \to \infty} \sqrt[n]{a^n+1}$ with $a \ge 0$ I'm  a bit rusty with limits
$$ \lim_{n \to \infty} \sqrt[n]{a^n+1}$$ with $a \ge 0$.
The solution in my book is $max \left \{ 0,1 \right \}$ but
my final results are:
1)  $+\infty$ if $0<a<1$
2)  $1$ if $a>1$
3)  $+\infty$ if $a=1$

*

*By Bernoulli's inequality$, 1 \le (1+a^n)^{1/n}\leq 1+\frac{a^n}{n} \to 1$ and so the limit is $1$ if $0 \le a < 1$.

*If $a>1$, then $a \le \sqrt[n]{a^n+1} \le \sqrt[n]{a^n+a^n} = a \sqrt[n]{2} \to a$ and so the limit is $a$.

*If $a=1$, then $\sqrt[n]{a^n+1}=\sqrt[n]{2} \to 1$.
You don't really need case 3 because the other two cases work for $a=1$.
Therefore,
$\displaystyle
\lim_{n \to \infty} \sqrt[n]{a^n+1} = \max(a,1)
$.
