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Function limits - Printable Version

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Function limits - elim - 10-13-2010 07:45 PM

(1) $\displaystyle{\lim_{x \to \infty} \frac{1}{\ln x} \int_0^{x}\frac{dt}{1+t^2\sin^2 t} = 1}$


RE: Function limits - elim - 12-17-2010 02:25 PM

(1) Let $\displaystyle{F(x) = \int_{0}^{x} \frac{dt}{1+t^2 \sin^2 t}}$, If $N$ is chosen so that $N\pi \leq x < (N+1)\pi$, then

$\begin{eqnarray*}\quad\quad F(x)
& = & F(N\pi) + O(1) = 2 \sum_{n=1}^{N} \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+n^2 \pi^2 \sin^2 x} + O(1) \\
& = & 2 \sum_{n=1}^{N} \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + (1+n^2 \pi^2) \tan^2 x} + O(1) = 2 \sum_{n=1}^{N} \int_{0}^{\infty} \frac{du}{1 + (1+n^2 \pi^2) u^2} + O(1) \\
& = & 2 \sum_{n=1}^{N} \frac{\pi}{2 \sqrt{1+n^2 \pi^2}} + O(1) = \sum_{n=1}^{N} \frac{1}{n \sqrt{1 + (n\pi)^{-2}}} + O(1) \\
& = & \sum_{n=1}^{N} \frac{1}{n} + O(1) = \ln x + O(1)
\end{eqnarray*}$

so we have $\displaystyle{\lim_{x\to\infty} \frac{1}{\ln x} \int_{0}^{x} \frac{dt}{1+t^2 \sin^2 t} = 1}$.