Calculate the limit $\lim_{n\to \infty }\int_{1}^{n}(n^2+x^3)^{-1/2} dx$ \begin{align}
\lim_{n\to \infty }\int_{1}^{n}\frac{dx}{\sqrt{n^2+x^3}}& \overset{x=nt}{=}\int_{1/n}^{1}\frac{dt}{\sqrt{1+nt^3}} \\
& \overset{t^3=p}{=}\int_{1/n^3}^{1}\frac{1/3p^{-2/3}}{\sqrt{1+np}}dp\\
& \overset{pn=q}{=}\frac{n^{-1/3}}{3}\int_{1/n^2}^{n}\frac{t^{-2/3}}{\sqrt{t+1}}dt \\
& \overset{t=\mathrm{tg}^2\theta }{=} 2\int_{0}^{\pi/2}\sin^{-1/3}\theta\cos^{-2/3}\theta d\theta\\
& =\mathrm{B}\left ( \frac{1}{3},\frac{1}{6} \right ) \\
& =\frac{\Gamma \left ( \frac{1}{3} \right )\Gamma \left ( \frac{1}{6} \right )}{\sqrt{\pi}}
\end{align}
Question: Have I calculated the limit correctly?
Why am I asking? After replacing $x=nt$, the limit and the integral can be interchanged and the limit will tend towards zero and the answer will be $0$.
Am I solving it correctly?
 A: Your mistake is saying
$$\frac{n^{-\frac{1}{3}}}{3}\int_{\frac{1}{n^{2}}}^{n}\frac{t^{-\frac{2}{3}}}{\sqrt{t+1}}dt = 2\int_{0}^{\frac{\pi}{2}}\sin\left(\theta\right)^{-\frac{1}{3}}\cos\left(\theta\right)^{-\frac{2}{3}}d\theta.$$
I'm not too sure how you tried using the substitution $t=tg^2\theta$ since that second integral does not depend on $n$. In fact, I'm not sure what $g$ is supposed to be.
The previous steps are correct, though, assuming you meant to keep writing $\displaystyle \lim_{n\to\infty}$ for each step.
(Answer) Going off $\frac{n^{-1/3}}{3}\int_{\frac{1}{n^{2}}}^{n}\frac{t^{-\frac{2}{3}}}{\sqrt{t+1}}dt$, an antiderivative of the integrand is $$\frac{n^{-1/3}}{3} \cdot 3t^{1/3} \cdot \left(_2F_1 \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-t\right)\right),$$ where we use the hypergeometric function. Using the Fundamental Theorem of Calculus and doing some simplifying, we get the integral to equal $$\frac{n^{-1/3}}{3}\left(\frac{3\left(-_{2}F_1 \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{n^2}\right)+n\left(_2F_1 \left(\frac{1}{3}, \frac{1}{2};\frac{4}{3};-n\right)\right)\right)}{n^{2/3}}\right),$$ where we assumed $n>0$. Applying $n\to\infty$, we get
$$
\eqalign{
&\displaystyle \lim_{n\to\infty} \frac{n^{-1/3}}{3}\left(\frac{3\left(-_{2}F_1 \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{n^2}\right)+n\left(_2F_1 \left(\frac{1}{3}, \frac{1}{2};\frac{4}{3};-n\right)\right)\right)}{n^{2/3}}\right) \cr
=& \left(\lim_{n\to\infty} \frac{n^{-1/3}}{3}\right)\left(\lim_{n\to\infty}\frac{3\left(-_{2}F_1 \left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{n^2}\right)+n\left(_2F_1 \left(\frac{1}{3}, \frac{1}{2};\frac{4}{3};-n\right)\right)\right)}{n^{2/3}}\right) \cr
=& 0 \cdot \frac{\Gamma{\left(1/6\right)\Gamma{\left(1/3\right)}}}{\sqrt{\pi}} \cr
=& 0. \cr
}
$$
A: I did direct substitution $x=n^{\frac{2}{3}}\tan^{\frac{2}{3}}\theta$ and $dx=\frac{2}{3}n^{\frac{2}{3}}\tan^{-\frac{1}{3}}\theta\sec^2\theta d\theta$ and got
$$\lim_{n\rightarrow\infty}\int_{\arctan(n^{-2/3})}^{\arctan(n^{1/3})} \frac{\frac{2}{3}n^{\frac{2}{3}}\tan^{-\frac{1}{3}}\sec^2\theta}{n\sec\theta}d\theta=(\lim_{n\rightarrow\infty}n^{-\frac{2}{3}})(\int_{0}^{\pi/2}\sin^{-\frac{1}{3}}\theta\cos^{-\frac{2}{3}}\theta d\theta)=0\times B(\frac{1}{3},\frac{1}{6})=0.$$
As usual first I checked with WA: https://www.wolframalpha.com/input?i=int_1%5E100000000+1%2Fsqrt%28100000000%5E2%2Bx%5E3%29dx
It couldn't solve the indefinite integral.
