Prove $\lim\limits_{n\to+\infty}\left(\frac{n^2-1}{n(1+n^2)}\right)^\frac{1}{\sqrt{n}}=1$ without $\lim_{x\to 0^+}x\ln(x)=0$ and continuity of $e^x$

Question

I am reading a real analysis book which in the chapter about sequences and series asks to prove that $\lim\limits_{n\to+\infty}\left(\frac{n^2-1}{n(1+n^2)}\right)^\frac{1}{\sqrt{n}}=1.$

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I have proved the statement by first noting that $\frac{n^2-1}{n(1+n^2)}=\frac{1}{n}\left(1-\frac{2}{n^2+1}\right)$ and since $1\geq\frac{1}{n}\left(1-\frac{2}{n^2+1}\right)\geq\frac{3}{5n}$ for every $n\geq 2$ it follows that $1\geq \left(\frac{n^2-1}{n(1+n^2)}\right)^\frac{1}{\sqrt{n}}\geq \left(\frac{3}{5n}\right)^\frac{1}{\sqrt{n}}$ for every $n\geq 2$ and since $\lim\limits_{x\to 0^+}x\ln(x)=0$ and $e^x$ is continuous at $x=0$ we also have $\lim\limits_{n\to+\infty}\left(\frac{3}{5n}\right)^\frac{1}{\sqrt{n}}=\lim\limits_{n\to+\infty}\exp\left(\frac{1}{\sqrt{n}}\ln\left(\frac{3}{5n}\right)\right)=\exp\left(\lim\limits_{n\to+\infty}\frac{1}{\sqrt{n}}\ln\left(\frac{3}{5n}\right)\right)=\exp(0)=1$ thus the claim follows by the squeeze theorem.

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Now, the thing that troubles me is that this problem is presented in the book way before talking about limits of functions and continuity so it stands to reason that there is a way to prove this result by only using the basic properties of functions, and the classic theorems about sequences and series but up to now I haven't been able to prove the statement this way so I would be interested in knowing how to do this. Thanks

Answer 1

For every integer $m\geqslant7,$ $m-3\geqslant\frac{m+1}2,$ whence $$ \binom{m}4=\frac{m(m-1)(m-2)(m-3)}{24}>\frac{(m+1)^4}{384}. $$ For every integer $n\geqslant49,$ $m\leqslant\sqrt{n}<m+1$ where $m\geqslant 7,$ therefore $$ \binom{m}4>\frac{n^2}{384}. $$ Let $a$ be any positive real number. Then for all $n\geqslant49,$ with $m=\left\lfloor\sqrt{n}\right\rfloor,$ $$ (1+a)^{\sqrt{n}}\geqslant(1+a)^m>\binom{m}4a^4>\frac{n^2a^4}{384}. $$ Therefore for all sufficiently large $n,$ $(1+a)^{\sqrt{n}}>n,$ whence $n^{1/\sqrt{n}}<1+a.$ Therefore $$ \lim_{n\to\infty}n^{1/\sqrt{n}}=1. $$

Also, for all $n>1,$ $$ \frac{n^2-1}{n^2+1}<\left(\frac{n^2-1}{n^2+1}\right)^{1/\sqrt{n}}<1, $$ and $$ \lim_{n\to\infty}\frac{n^2-1}{n^2+1}=1, $$ therefore $$ \lim_{n\to\infty}\left(\frac{n^2-1}{n^2+1}\right)^{1/\sqrt{n}}=1. $$ Taking the quotient of these two limits, $$ \lim_{n\to\infty}\left(\frac{n^2-1}{n(n^2+1)}\right)^{1/\sqrt{n}}=1. $$

Answer 2

Hint: You can show $\displaystyle\left(\frac{n^2-1}{n(1+n^2)}\right)^\frac{1}{\sqrt{n}}=\displaystyle\left(\left(1+1\frac{1}{\frac{n^3+1}{-n^3+n^2-n-1}}\right)^{\frac{n^3+1}{-n^3+n^2-n-1}}\right)^{\frac{1}{\sqrt n}\cdot\frac{n^3+1}{-n^3+n^2-n-1}}$.
From here you can use the fact that $\displaystyle\lim_{n\to\infty}{\left(1+\frac{1}{n}\right)^{n}}=e$.