Ermittle das Verhalten der Funktionswerte von $ff$ an den Rändern von $D_{f}D\_f$

Betrachtung des linken Randes des Definitionsbereichs: ${\displaystyle \underset{x\to 0^{+}}{\lim }f(x)}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow0^+\}\; f(x)$

Rechtsseitiger Grenzwert: ${\displaystyle \underset{x\to 0^{+}}{\lim }(\frac{\pi }{4}-\arctan \left(\begin{array}{c}{\displaystyle \frac{1-x}{x}}\end{array}\right))}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow0^+\}\backslash left(\backslash dfrac\{\backslash pi\}\{4\}-\backslash arctan\backslash begin\{pmatrix\}\backslash dfrac\{1-x\}\{x\}\backslash end\{pmatrix\}\backslash right)$

Dabei gilt: ${\displaystyle \underset{x\to 0^{+}}{\lim }\frac{1-x}{x}=\mathrm{\infty }}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow0^+\}\backslash dfrac\{1-x\}\{x\}=\backslash infty$ und ${\displaystyle \underset{x\to \mathrm{\infty }}{\lim }\arctan \left(\begin{array}{c}{\displaystyle \frac{1-x}{x}}\end{array}\right)=\frac{\pi }{2}}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow\backslash infty\}\backslash arctan\backslash begin\{pmatrix\}\backslash dfrac\{1-x\}\{x\}\backslash end\{pmatrix\}=\backslash dfrac\{\backslash pi\}\{2\}$ (siehe Abbildung).

Dann erhält man: ${\displaystyle \underset{x\to 0^{+}}{\lim }(\frac{\pi }{4}-\arctan \left(\begin{array}{c}{\displaystyle \frac{1-x}{x}}\end{array}\right))=\frac{\pi }{4}-\frac{\pi }{2}=-\frac{\pi }{4}}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow0^+\}\backslash left(\backslash dfrac\{\backslash pi\}\{4\}-\backslash arctan\backslash begin\{pmatrix\}\backslash dfrac\{1-x\}\{x\}\backslash end\{pmatrix\}\backslash right)=\backslash dfrac\{\backslash pi\}\{4\}-\backslash dfrac\{\backslash pi\}\{2\}=-\backslash dfrac\{\backslash pi\}\{4\}$

Betrachtung des rechten Randes des Definitionsbereichs: ${\displaystyle \underset{x\to \mathrm{\infty }}{\lim }f(x)}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow\backslash infty\}f(x)$

Dabei gilt: ${\displaystyle \underset{x\to \mathrm{\infty }}{\lim }\frac{1-x}{x}=-1}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow\backslash infty\}\backslash dfrac\{1-x\}\{x\}=-1$ und ${\displaystyle \underset{x\to \mathrm{\infty }}{\lim }\arctan \left(\frac{1-x}{x}\right)=-\frac{\pi }{4}}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow\backslash infty\}\backslash arctan\backslash left(\backslash dfrac\{1-x\}\{x\}\backslash right)=-\backslash dfrac\{\backslash pi\}\{4\}$ (siehe Abbildung).

Dann erhält man: ${\displaystyle \underset{x\to \mathrm{\infty }}{\lim }(\frac{\pi }{4}-\arctan \left(\begin{array}{c}{\displaystyle \frac{1-x}{x}}\end{array}\right))=\frac{\pi }{4}-(-\frac{\pi }{4})=\frac{\pi }{2}}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow\backslash infty\}\backslash left(\backslash dfrac\{\backslash pi\}\{4\}-\backslash arctan\backslash begin\{pmatrix\}\backslash dfrac\{1-x\}\{x\}\backslash end\{pmatrix\}\backslash right)=\backslash dfrac\{\backslash pi\}\{4\}-\backslash left(-\backslash dfrac\{\backslash pi\}\{4\}\backslash right)=\backslash dfrac\{\backslash pi\}\{2\}$

Ergebnis:

Verhalten am linken Rand von $D_f:f(x)\mapsto -{\displaystyle \frac{\pi }{4}}D\_f:\; f(x)\backslash mapsto\; -\backslash dfrac\{\backslash pi\}\{4\}$

Verhalten am rechten Rand von $D_f:f(x)\mapsto {\displaystyle \frac{\pi }{2}}D\_f:\; f(x)\backslash mapsto\; \backslash dfrac\{\backslash pi\}\{2\}$

Berechne die Nullstelle von $ff$

Es ist $f(x)={\displaystyle \frac{\pi }{4}}-\arctan \left(\begin{array}{c}{\displaystyle \frac{1-x}{x}}\end{array}\right)f(x)=\backslash dfrac\{\backslash pi\}\{4\}-\backslash arctan\backslash begin\{pmatrix\}\backslash dfrac\{1-x\}\{x\}\backslash end\{pmatrix\}$.

Für die Nullstellen löse die Gleichung $f(x)=0f(x)=0$.

$0={\displaystyle \frac{\pi }{4}}-\arctan \left(\begin{array}{c}{\displaystyle \frac{1-x}{x}}\end{array}\right)\Rightarrow {\displaystyle \frac{\pi }{4}}=\arctan \left(\begin{array}{c}{\displaystyle \frac{1-x}{x}}\end{array}\right)0=\backslash dfrac\{\backslash pi\}\{4\}-\backslash arctan\backslash begin\{pmatrix\}\backslash dfrac\{1-x\}\{x\}\backslash end\{pmatrix\}\backslash ;\backslash Rightarrow\backslash ;\backslash dfrac\{\backslash pi\}\{4\}=\backslash arctan\backslash begin\{pmatrix\}\backslash dfrac\{1-x\}\{x\}\backslash end\{pmatrix\}$

$\arctan (1)={\displaystyle \frac{\pi }{4}}\Rightarrow {\displaystyle \frac{1-x}{x}}=1\Rightarrow 1-x=x\Rightarrow x=\mathrm{0,5}\backslash arctan(1)=\backslash dfrac\{\backslash pi\}\{4\}\backslash ;\backslash Rightarrow\backslash ;\backslash dfrac\{1-x\}\{x\}=1\backslash ;\backslash Rightarrow\backslash ;1-x=x\backslash ;\backslash Rightarrow\backslash ;x=0\{,\}5$

Die Nullstelle von $ff$ ist $x=\mathrm{0,5}x=0\{,\}5$.