Question P.173: Find lim n→∞ ∫0^∞ n cos( √^4 x/n²)/1 + n²x²dx. Idea. First, ......
Find \lim_{n→\infty} \frac{\int_0^{\infty} n \cos( \sqrt[4]{ x/n^2})}{1 + n^2x^2}dx.
Idea. First, we note that we cannot switch the limit and the integral, for if we could, we could start by showing that \lim_{n→\infty}\frac{ n \cos(\sqrt[4]{ x/n^2})}{1 + n^2x^2} = 0 for any nonzero x, and then conclude that the answer would be 0. This answer is incorrect, and a bit of numerical approximation using technology yields approximate values of the integral as follows:
n approximate value
3 1.37144
5 1.47507
200 1.5704
2000 1.57078
These values suggest that the limit might be π/2 ≈ 1.5708. To prove this, we note that once n gets large the integrand will be close to zero unless x is small. On the other hand, for values of x reasonably close to zero, \sqrt[4]{ x/n^2} should be close to zero, so \cos (\sqrt[4]{ x/n^2}) should be close to 1. Therefore, we can hope that the integral is close to \left.\int_0^{\infty} \frac{n dx}{1 + n^2x^2} = \arctan(nx)\right|_0^{\infty}= π/2.
With the reasoning just presented, we see that it will be enough to show that
\lim_{n\rightarrow\infty}\left(\int_{0}^{\infty}{\frac{n\,d x}{1+n^{2}x^{2}}}-\int_{0}^{\infty}\,{\frac{\,\,n\,\cos\left({\sqrt[4]{x/n^{2}}}\,\right)\,}{1+n^{2}x^{2}}}\,d x\right)=0,so we want to show that
\lim_{n\rightarrow\infty}\int_{0}^{\infty}\ {\frac{n\left(1-\cos\left({\frac{\sqrt[4]{x/n^{2}}}{1+n^{2}x^{2}}}\right)\right)}{1+n^{2}x^{2}}}\,d x\,=\,0..For this, note that
0\leq\int_{0}^{\infty}~\frac{n\left(1-\cos\left(\sqrt[4]{x/n^{2}}\right)\right)}{1+n^{2}x^{2}}\,d x=\int_{0}^{\infty}~\frac{2n\sin^{2}\left(\frac{1}{2}\,{\sqrt[4]{x/n^{2}}}\right)}{1+n^{2}x^{2}}\,d x\\\lt \int_{0}^{\infty}\frac{\frac{1}{2}\sqrt{x}}{1+n^{2}x^{2}}\,d x\qquad (because~\sin \theta\lt \theta~~for~\theta\gt 0 )\\={\frac{1}{2n^{3/2}}}\int_{0}^{\infty}{\frac{\sqrt{u}}{1+u^{2}}}\,d u\qquad{\mathrm{(with~}}u=n x).Because the last integral converges to a value independent of n, the result follows by the squeeze principle.