$ \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 $a=1$
 A: 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.
A: *

*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)
$.
A: 1) $0<a<1:$
$1 \lt (a^n +1)^{1/n} \lt (2)^{1/n} .$
$1\le \lim_{n \rightarrow \infty} (a^n+1)^{1/n} \le $
$\lim_{ n\rightarrow \infty }2^{1/n} =1.$
2) $a>1:$
$a \lt (a^n +1)^{1/n} \lt (2a^n)^{1/n}.$
$a \le \lim_{ n \rightarrow \infty}(a^n+1)^{1/n} \le $
$\lim_{n \rightarrow \infty}(2)^{1/n}a = a.$
3)$a=1. $
Can you do it?
A: I agree with
$\sqrt[n]{a^n+1}=e^{log\sqrt[n]{a^n+1} }=e^{\frac{log(a^n+1)}{n} }$
If $a>1$  $\frac{log(a^n+1)}{n}<\frac{log(a^n)}{n}=\frac{n*log(a)}{n}=log(a)$ and then $lim_n \sqrt[n]{a^n+1}=e^{log {a}}=a$
If $0<a<1$  $\frac{log(a^n+1)}{n}<\frac{log(a^n)}{n}=\frac{n*log(a)}{n}=log(a)$ and then $lim_n \sqrt[n]{a^n+1}=e^{log {a}}=a$   (but $a<1$!!!)
If $a=1$ ,   $\sqrt[n]{a^n+1}=\sqrt[n]{1^n+1} $and then  $lim_n \sqrt[n]{1^n+1}=lim_n \sqrt[n]{2}=1$
If $a=0$ ,   $\sqrt[n]{a^n+1}=\sqrt[n]{1} $  and then  $lim_n \sqrt[n]{1}=1$
I'm not sure about the case of $0<a<1$ because a <1 and the result is not the maximum between a and 1.
