Calculate $\lim_{n \to \infty } \frac{1}{n^2} \int_0^n \frac{ \sqrt{n^2-x^2} }{2+x^{-x}} dx$

Question

I need to calculate the limit$$\lim_{n \to \infty } \frac{1}{n^2} \int_0^n \frac{ \sqrt{n^2-x^2} }{2+x^{-x}} dx$$ How could I calculate this? Any hlep would be appreciated.

Answer 1

Outline:

• First, do the change of variable $x=ny$, to get $ \int_0^1 f_n(y) dy $ with $$f_n(y) = \frac{\sqrt{1-y^2}}{2+e^{-ny \ln (ny)}}.$$

• Then, compute the pointwise limit $f$ of $f_n$ on $(0,1]$.

• Finally, apply the Dominated Convergence Theorem to show that $\int_0^1 f_n \xrightarrow[n\to\infty]{} \int_0^1 f$.

Answer 2

Try $u=x/n$. It'll get rid of most of the horrible places where $n$ appears. All that's left to do is figure out the limit in the denominator, where you get

$$\lim_{n\to\infty}(nu)^{-nu}$$