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.

$$\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2=\lim\limits_{x \to \infty}x^2\left[ e^{\frac{1}{n}\ln \left(1+ \frac{1}{x^2} \right)\cdots \left(1+ \frac{n}{x^2} \right)}-1 \right] =\\=\lim\limits_{x \to \infty}\frac{1}{n}x^2\left[  \ln \left(1+ \frac{1}{x^2} \right)+\cdots+\ln \left(1+ \frac{n}{x^2} \right)\right]=\frac{1}{n}\left[1+2+ \cdots+n\right] =\frac{n+1}{2} $$
