Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $ I'm trying to calculate:
$$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$
Here is my attempt.
Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become
\begin{align*}
T &= \lim\limits_{t \to 0} \sqrt[n]{\left(1+\dfrac{1}{t}\right)\left(2+\dfrac{1}{t}\right)...\left(n+\dfrac{1}{t}\right)}-\dfrac{1}{t}\\
&=\lim\limits_{t \to 0} \sqrt[n]{\left(\dfrac{t+1}{t}\right)\left(\dfrac{2t+1}{t}\right)...\left(\dfrac{nt+1}{t}\right)}-\dfrac{1}{t} \\
&=\lim\limits_{t \to 0} \dfrac{\sqrt[n]{(t+1)(2t+1)...(nt+1)}-1}{t}
\end{align*}
My idea is to use $\lim\limits_{x\to0}\dfrac{(ax+1)^{\beta}-1}{x} =a\beta .$ But after some steps (above), now I'm stuck.
Thanks for any helps.
 A: The idea is very good! The limit should be for $t\to0^+$, but since the limit for $t\to0$ exists, there's no real problem. However, you should use $t\to0^+$ for the sake of rigor.
The two-sided limit is the derivative at $0$ of the function
$$
f(t)=\sqrt[n]{(t+1)(2t+1)\dotsm(nt+1)}
$$
and in order to compute it, the logarithmic derivative is handy:
$$
\log f(t)=\dfrac{1}{n}\bigl(\log(t+1)+\log(2t+1)+\dots+\log(nt+1)\bigr)
$$
and therefore
$$
n\frac{f'(t)}{f(t)}=\frac{1}{t+1}+\frac{2}{2t+1}+\dots+\frac{n}{nt+1}
$$
which yields
$$
n\frac{f'(0)}{f(0)}=1+2+\dots+n=\frac{n(n+1)}{2}
$$
Since $f(0)=1$, we have
$$
f'(0)=\frac{n+1}{2}
$$
A: Based on your way, in fact you can calculate $(1+t)(1+2t)...(1+nt)$ directly.
$$(1+t)(1+2t)...(1+nt)=a_nt^n+...+\frac{(n+1)n}{2}t+1=f(t)+1$$
and $f(t)$ tends to $0$.
So
$$(f(t)+1)^{\frac{1}{n}}-1\sim\frac{1}{n}f(t)$$
and
$$T=\lim_{t\rightarrow 0^+}\frac{1}{n}\frac{f(t)}{t}=\frac{n+1}{2}$$
A: The limit can also be shown using HM-GM-AM.
Setting $u = x^2$ and considering $u\to +\infty$ we have
$$\frac n{\sum_{k=1}^n\frac 1{k+u}} - u \leq \sqrt[n]{\prod_{k=1}^n (k+u)} - u \leq \frac{\sum_{k=1}^n(k+u)}n - u = \frac{n+1}{2}$$
For the LHS we have
$$\frac n{\sum_{k=1}^n\frac 1{k+u}} - u = \frac{n - \sum_{k=1}^n\frac u{k+u}}{\frac 1u\sum_{k=1}^n\frac 1{\frac ku+1}} $$ $$=  \frac{u \sum_{k=1}^n\frac k{k+u}}{\sum_{k=1}^n\frac 1{\frac ku+1}}= \frac{\sum_{k=1}^n\frac k{\frac ku+1}}{\sum_{k=1}^n\frac 1{\frac ku+1}}\stackrel{u\to+\infty}{\longrightarrow}\frac{\sum_{k=1}^n k}{n} = \frac{n+1}{2}$$
Now, squeezing gives the limit $ \frac{n+1}{2}$.
A: By definition of pochammer symbol  $$(x^2+1)^{(n)}=(1+x^2)(2+x^2)\cdots (n+x^2)=\frac{\Gamma(x^2+1+n)}{\Gamma(x^2+1)}\sim x^{2n}\left(1+\frac{n(n+1)}{2x^2}+O(x^{-4})\right)$$ thus $$\sqrt[n]{(x^2+1)^{(n)}}-x^2 =x^2\left(1+\frac{n(n+1)}{2x^2}\right)^{\frac{1}{n}}-x^2$$ using the fractional binomial theorem  we have limit $$\lim_{x\to \infty}\left(\sqrt[n]{(x^2+1)^{(n)}}-x^2\right)= x^2\left(1+\frac{n(n+1)}{2n} x^{-2} +O(x^{-4}) -x^2\right)=\frac{n+1}{2}$$
Notation: $O(.)$ is Big O  notation.
A: Hint:
In the development of the product under the radical, the dominant terms are
$$x^{2n}+(1+2+\cdots n)x^{2n-2}=x^{2n}\left(1+\frac{n(n+1)}2x^{-2}\right).$$
Then, taking the $n^{th}$ root, ($\sqrt[n]{1+x}=1+\frac xn+\cdots$),
$$x^2\left(1+\frac{n(n+1)}{2n}x^{-2}+\cdots\right)-x^2\to\frac{n+1}2$$ as the other terms are of a lower order.
You can make it rigorous with the $o$ notation.
A: $$\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2=\lim\limits_{x \to \infty}x^2\left[ e^{\frac{1}{n}\ln \left(1+ \frac{1}{x^2} \right)\cdots \left(1+ \frac{n}{x^2} \right)}-1 \right] =\\=\lim\limits_{x \to \infty}\frac{1}{n}x^2\left[  \ln \left(1+ \frac{1}{x^2} \right)+\cdots+\ln \left(1+ \frac{n}{x^2} \right)\right]=\frac{1}{n}\left[1+2+ \cdots+n\right] =\frac{n+1}{2} $$
