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