Show that $\lim\limits_{n\to\infty}\sqrt [n] {n^{2}+n+1}=1$

Question

Proof. Note that $1\leq \sqrt [n] {n^2+n+1}\leq \sqrt [n] {3n^{2}}$. But, $\lim\limits_{n\to\infty}\sqrt [n] {3n^{2}} =1$. Therefore, by the squeeze lemma, we have $\lim\limits_{n\to\infty} \sqrt [n]{n^2+n+1} =1$

My question is: How did $\lim\limits_{n\to\infty} \sqrt [n] {3n^2} =1$ be? So, how can it prove as primitive methods? We cannot use L'hopital rule, rules of derivative, exp functions.

Answer 1

Hint let $l=n^{2/n} $ so $\log (l)=\frac {2}{n}\log (n) $ now we know $n $ grows faster than $log (n) $ so $log (l)=0$ thus $l=e^0=1$ thus limit is $1$

Answer 2

We can write $\sqrt[n]{3n^2} = (\sqrt{3}n)^\frac{2}{n} = e^\frac{2\ln(\sqrt{3}n)}{n}$. Note that for $n > 1$ we have $0 < \ln(n) < n$. If $n > 1$ then $\sqrt{\sqrt{3}n} > 1$ (since $\sqrt{3}$ > 1). Thus we have for $n > 1$ \begin{align} 0 < \ln(\sqrt{\sqrt{3}n}) < \sqrt{\sqrt{3}n}\\ 0 < 2\ln(\sqrt{\sqrt{3}n}) < 2\sqrt{\sqrt{3}n}\\ 0 < \ln(\sqrt{3}n) < 2\sqrt{\sqrt{3}n}\\ 0 < \frac{\ln(\sqrt{3}n)}{n} <\frac{2\sqrt{\sqrt{3}} \sqrt{n}}{\sqrt{n} \sqrt{n}} = \frac{2\sqrt[4]{3}}{\sqrt{n}} \end{align} Since $\lim_{n \to \infty} \frac{2\sqrt[4]{3}}{\sqrt{n}} = 0$ we have by the squeeze theorem $\lim_{n \to \infty} \frac{\ln(\sqrt{3}n)}{n} = 0$. Thus we have \begin{equation} \lim_{n\to\infty} \sqrt[n]{3n^2} = \lim_{n \to \infty} e^\frac{2\ln(\sqrt{3}n)}{n} = e^{2\lim_{n\to\infty} \frac{\ln(\sqrt{3}n)}{n}} = e^{2(0)} = e^0 = 1 \end{equation}

Answer 3

Write $n^{1/n} = 1 + a_n.$ We have $0<a_n$ for all $n.$ The binomial formula shows

$$n = (1 + a_n)^n = 1 + na_n + [n(n-1)/2]a_n^2 + \cdots$$

for $n>1.$ Thus for $n>1$ we have $n>[n(n-1)/2]a_n^2,$ which implies $0 < a_n < \sqrt {2/(n-1)}.$ It follows that $a_n \to 0,$ giving $n^{1/n} \to 1.$

Answer 4

Let $y = (3n^2)^{\frac 1n}$. Then $\ln y = \dfrac{\ln(3n^2)}{n}$. Note $ \dfrac{d}{dn}ln(3n^2) = \dfrac{6n}{3n^2}=\dfrac 2n$. So, by L'Hospital's rule,$\displaystyle \lim_{n \to \infty} \ln y = \lim_{n \to \infty}\dfrac{\ln(3n^2)}{n} = \lim_{n \to \infty}\dfrac{2}{n^2} = 0$. Because $\ln$ is continuous, this implies that $\displaystyle \ln\left(\lim_{n \to \infty}y\right) = 0$. Hence $\displaystyle \lim_{n \to \infty}y = e^0 = 1.$

Answer 5

see my nice answer: $$\lim_{n\to \infty}n^{1/n}=e^{\lim_{n\to \infty}\large \frac1n\ln n}$$ apply l'hopital's rule, $$=e^{\lim_{n\to \infty}\large \frac{1/n}{1}}=e^{\lim_{n\to \infty}\large \frac1n}=1$$

Answer 6

Actually, we have a more general result, that is $$ \lim _{n \to \infty} \sqrt[n]{n} = 1 , $$ because the latter limit deduces $$ \lim _{n \to \infty} \sqrt[n]{3n^2} = \lim _{n \to \infty} \sqrt[n]{3} \cdot ( \sqrt[n]{n} )^2 = 1\cdot 1^2 = 1 . $$ Now we prove the first limit using the deffinition of limit, i.e. prove: $$ \forall \epsilon > 0, ~ \exists N_\epsilon > 0, ~~ \textrm{such that} ~~ \forall n > N_\epsilon, ~~ \left| \sqrt[n]{n} - 1 \right| < \epsilon $$ Let $\epsilon \in (0,1) $ arbitrarily, choose $N_\epsilon$ to be large enough ($N_\epsilon$ depends on $\epsilon$) such that $$ (1-\epsilon)^{N_\epsilon} < N_\epsilon < (1+\epsilon)^{N_\epsilon} $$ This is a quite simple task since $0< 1-\epsilon < 1$ and $1 + \epsilon > 1$, and once we make a correct choice of $N _ \epsilon$, the above inequalities hold for any $n > N_\epsilon$, thus, $ \left| \sqrt[n]{n} - 1 \right| < \epsilon $ for any $n > N_\epsilon$. Game over.