Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $

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 $$ f'(0)=\frac{n+1}{2} $$

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 $$T=\lim_{t\rightarrow 0^+}\frac{1}{n}\frac{f(t)}{t}=\frac{n+1}{2}$$

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}$.