$x_n$ is the $n$'th positive solution to $x=\tan(x)$. Find $\lim_{n\to\infty}\left(x_n-x_{n-1}\right)$.
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user72273
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Do you see the solution, intuitively ? – leonbloy Apr 17 '13 at 19:19
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2It certainly converges, since there is a root in every interval $[ n\pi,(n+1)\pi]$. Thus $x_{n}-x_{n-1}<\pi$ for all $n$. It seems that the limit is $\pi$, but I can't think how to show it rigorously. – preferred_anon Apr 17 '13 at 19:22
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@GitGud Whoops. – user72273 Apr 17 '13 at 19:23
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See this answer: http://math.stackexchange.com/a/18733/312 – leonbloy Apr 17 '13 at 19:26
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HINT
Show that $n \pi < x_n < n \pi + \dfrac{\pi}2$, where $x_0 = 0$ by considering the function $f(x) = x - \tan(x)$, which is continuous in the interval $n \pi < x_n < n \pi + \dfrac{\pi}2$.
Now consider $y_n = n \pi + \dfrac{\pi}2 - x_n$. We have $y_n \in (0,\pi/2)$. Since $x_n = \tan(x_n)$, we have $$n \pi + \dfrac{\pi}2 - y_n = \cot(y_n) \in \left(n \pi, n \pi + \dfrac{\pi}2\right) \implies \underbrace{\tan(y_n) \in \left(\dfrac1{n \pi + \dfrac{\pi}2}, \dfrac1{n \pi}\right) \implies y_n < \dfrac1{n \pi}}_{\text{Because $0 < y_n < \tan(y_n)$}}$$ Hence, we have $$\left \vert x_n - \left(n \pi + \dfrac{\pi}2\right) \right \vert < \dfrac1{n \pi}$$ Conclude what you want from this.
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HINT: Sketch $x = \mathrm{tan}(x)$ and think about the asymptotes of tan$(x)$.
user27182
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