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.

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}$.
