Find lim n→∞ ∫0^∞ n cos( √^4 x/n²)/1 + n²x²dx. 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→∞ n cos(√^4 x/n²)/1 + n²x² = 0 for any nonzero x, and then conclude that the answer would be 0. This answer is incorrect, and a

Question

Find [latex]\lim_{n→\infty} \frac{\int_0^{\infty} n \cos( \sqrt[4]{ x/n^2})}{1 + n^2x^2}dx.[/latex]
Idea. First, we note that we cannot switch the limit and the integral, for if we could, we could start by showing that [latex]\lim_{n→\infty}\frac{ n \cos(\sqrt[4]{ x/n^2})}{1 + n^2x^2} = 0[/latex] 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, [latex]\sqrt[4]{ x/n^2}[/latex] should be close to zero, so [latex]\cos (\sqrt[4]{ x/n^2}) [/latex] should be close to 1. Therefore, we can hope that the integral is close to [latex]\left.\int_0^{\infty} \frac{n dx}{1 + n^2x^2} = \arctan(nx)\right|_0^{\infty}= π/2.[/latex]

Accepted Answer

With the reasoning just presented, we see that it will be enough to show that

[latex]\lim_{n ightarrow\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}}}\, ight)\,}{1+n^{2}x^{2}}}\,d x ight)=0,[/latex]

so we want to show that

[latex]\lim_{n ightarrow\infty}\int_{0}^{\infty}\ {\frac{n\left(1-\cos\left({\frac{\sqrt[4]{x/n^{2}}}{1+n^{2}x^{2}}} ight) ight)}{1+n^{2}x^{2}}}\,d x\,=\,0.[/latex]

.For this, note that

[latex]0\leq\int_{0}^{\infty}~\frac{n\left(1-\cos\left(\sqrt[4]{x/n^{2}} ight) ight)}{1+n^{2}x^{2}}\,d x=\int_{0}^{\infty}~\frac{2n\sin^{2}\left(\frac{1}{2}\,{\sqrt[4]{x/n^{2}}} ight)}{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 heta\lt heta~~for~ heta\gt 0 )\={\frac{1}{2n^{3/2}}}\int_{0}^{\infty}{\frac{\sqrt{u}}{1+u^{2}}}\,d u\qquad{\mathrm{(with~}}u=n x).[/latex]

Because the last integral converges to a value independent of n, the result follows by the squeeze principle.

Question P.173: Find lim n→∞ ∫0^∞ n cos( √^4 x/n²)/1 + n²x²dx. Idea. First, ......

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