Find this limit, When n to infinity I don´t understand this problem, I need help with this
Find the next limit. Justify your procedure
$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c} $$
I Tried do this, but I think it's wrong
$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c} = \displaystyle \lim_{n \to \infty} {n + c}^n $$
After that, I tried use a stolz criterium $$\displaystyle \lim_{n \to \infty} \sqrt[b_{n}]{a_{n}} = \displaystyle \lim_{n \to \infty} \sqrt[b_{n+1}-b_{n}]{\dfrac{a_{n+1}}{a_{n}}}$$
where $$a_{n} = n+c , b_{n} =n , a_{n+1} = n+c+1 , b_{n+1} = n+1 , \displaystyle \lim_{n \to \infty} b_{n} = \displaystyle \lim_{n \to \infty} n = \infty$$
so $$\displaystyle \lim_{n \to \infty} \sqrt[n]{n+c} = \displaystyle \lim_{n \to \infty} \sqrt[n+1-n]{\dfrac{n+c+1}{n+c}} = \displaystyle \lim_{n \to \infty} \sqrt[1]{\dfrac{n+c+1}{n+c}} = \displaystyle \lim_{n \to \infty} \dfrac{n+c+1}{n+c} = \dfrac{\displaystyle \lim_{n \to \infty} n+c+1 }{\displaystyle \lim_{n \to \infty} n+c }$$
I got this but I don't know if this is right or not,
 A: Using the formula $e^{ln(x)}=x$:
$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c}=\lim_{n \to \infty} (n + c)^{\frac{1}{n}}= e^{\lim_{n \to \infty}\frac{ln(n + c)}{n}}= e^{\lim_{n \to \infty}\frac{1}{n+c}}=e^0=1$$
If you had persisted with what you were trying:
$$\dfrac{\displaystyle \lim_{n \to \infty} n+c+1 }{\displaystyle \lim_{n \to \infty} n+c } = \displaystyle \lim_{n \to \infty}\dfrac{ n+c+1 }{ n+c } = \displaystyle \lim_{n \to \infty}(1+\frac{1}{n+c})=1+0=1$$
A: Here is yet another solution  taking $\lim_{n\rightarrow\infty}\sqrt[n]{n}=1$ as a known fact.
For all $n> |c|$ we have that
$$ 1+|c|+c\leq n+c\leq 2n$$
Hence
$$ (1+|c|+ c)^{1/n}\leq (n+c)^{1/n}\leq 2^{1/n}n^{1/n}$$
the conclusion follows from the squeeze lemma and the fact that for any $a>0$, $\lim_{n\rightarrow\infty}\sqrt[n]{a}=1$.

About your attempt:
There is a well known result that says that for any sequence $\{a_n:n\in\mathbb{N}\}\subset\mathbb{C}$
$$
\liminf_{n\rightarrow\infty}\sqrt[n]{|a_n|}\leq\liminf_{n\rightarrow\infty}\frac{|a_{n+1}|}{|a_n|}\leq \limsup_{n\rightarrow\infty}\frac{|a_{n+1}|}{|a_n|}\leq \limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}
$$
That is closer to the argument you try to use.
A: By Bernoulli's inequality,
if $x \ge -n$ then
$(1+x/n)^n \ge 1+x
$
so
$(1+x)^{1/n}
\le 1+x/n$.
Therefore,
for $n \ge |c|$,
$\begin{array}\\
(n+c)^{1/n}
&=n^{1/n}(1+c/n)^{1/n}\\
&\le n^{1/n}(1+c/n^2)\\
&\le n^{1/n}+cn^{1/n}/n^2\\
&= n^{1/n}+2c/n^2
\qquad\text{since } 2^n \ge n \text{ or } n^{1/n} \le 2\\
\end{array}
$
Since $n^{1/n} \to 1$
and $2c/n^2 \to 0$,
$(n+c)^{1/n} \to 1$.
A: Another way to do it.
$$a_n= \sqrt[n]{n + c}\implies\log(a_n)=\frac 1 n \log(n+c)=\frac 1 n\left( \log \left(1+\frac{c}{n}\right)+\log (n)\right)$$ Since $n$ is large
$$\log \left(1+\frac{c}{n}\right)\sim \frac{c}{n}\implies
\log(a_n)\sim \frac {\log(n)}n+\frac{c}{n^2}\to 0\implies a_n \to 1$$
