0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.