Find this limit, When n to infinity

Question

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,

Answer 1

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: $$\dfrac{\displaystyle \lim_{n \to \infty} n+c+1 }{\displaystyle \lim_{n \to \infty} n+c } = \displaystyle \lim_{n \to \infty}\dfrac{ n+c+1 }{ n+c } = \displaystyle \lim_{n \to \infty}(1+\frac{1}{n+c})=1+0=1$$

Answer 2

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 $2c/n^2 \to 0$, $(n+c)^{1/n} \to 1$.

Answer 3

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

$$\log \left(1+\frac{c}{n}\right)\sim \frac{c}{n}\implies \log(a_n)\sim \frac {\log(n)}n+\frac{c}{n^2}\to 0\implies a_n \to 1$$

Answer 4

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 $a>0$, $\lim_{n\rightarrow\infty}\sqrt[n]{a}=1$.

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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.