# Find $\lim_{x\rightarrow{\infty}} \frac{\int_{0}^{1} (1+(xy)^{n})^{\frac{1}{n}}dy}{x}$, where $n\geq 2$.

Find $$\lim_{x\rightarrow{\infty}} \frac{\int_{0}^{1} (1+(xy)^{n})^{\frac{1}{n}}dy}{x}$$ where $$n\geq 2$$.

I introduced the $$x$$ in the integral and also, the limit and so I obtained it is equal to limit when $$x$$ tends to $$\infty$$ and I obtained integral from $$y$$ over $$[0,1]$$, so $$\frac{1}{2}$$. I am not sure it's correct.

Yes, $$1/2$$ is the correct answer for $$n\geq 1$$. Let $$t=xy$$ then $$\int_{0}^{1} (1+(xy)^{n})^{\frac{1}{n}}dy=\frac{1}{x}\int_{0}^{x} (1+t^{n})^{\frac{1}{n}}dt$$ Hence the given limit is $$\lim_{x\rightarrow{+\infty}} \frac{\int_{0}^{1} (1+(xy)^{n})^{\frac{1}{n}}dy}{x} =\lim_{x\rightarrow{+\infty}} \frac{\int_{0}^{x} (1+t^{n})^{\frac{1}{n}}dt}{x^2} =\lim_{x\rightarrow{+\infty}} \frac{ (1+x^{n})^{\frac{1}{n}}}{2x} \\=\lim_{x\rightarrow{+\infty}} \frac{ (\frac{1}{x^{n}}+1)^{\frac{1}{n}}}{2}=\frac{1}{2}$$ where at the second step we used L'Hopital theorem.