$ \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 $a=1$

The correct result is $\max\{a, 1\}$, so I suppose it's a typo or a miscopied expression. To see this, note that


*

*If $a \le 1$, then $1 \le a^n + 1 \le 2$ for all $n$. Taking $n$-th roots and a limit gives limit $1$.

*If $a > 1$, then
$$\left(a^n + 1\right)^{1/n} = a \left(1 + \frac{1}{a^n}\right)^{1/n}$$
Now apply the previous case with $1/a$ to see why this tends to $a \cdot 1 = a$.

The key idea here is that if $a < 1$, the term $a^n$ is negligible once $n$ is large. If $a > 1$, the term $1$ is negligible once $n$ is large.
