«K12/K13» 83. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 83. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 115, Aufgabe 61

Berechne:

a)

\lim\limits_{x \to \infty} \left(\frac{2x + 1}{2x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(\frac{2x - 1 + 1 + 1}{2x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(1 + \frac{2}{2x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(1 + \frac{2}{2x - 1}\right)^{2x - 1 + 1} = \lim\limits_{u \to \infty} \left(1 + \frac{2}{u}\right)^{u + 1} = \lim\limits_{u \to \infty} \left(1 + \frac{2}{u}\right)^u \cdot \left(1 + \frac{2}{u}\right) = e^2 \cdot 1 = e^2;$\underset{x\to \infty }{lim}{\left(\frac{2x+1}{2x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(\frac{2x-1+1+1}{2x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(1+\frac{2}{2x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(1+\frac{2}{2x-1}\right)}^{2x-1+1}=\underset{u\to \infty }{lim}{\left(1+\frac{2}{u}\right)}^{u+1}=\underset{u\to \infty }{lim}{\left(1+\frac{2}{u}\right)}^u\cdot \left(1+\frac{2}{u}\right)=e^2\cdot 1=e^2;$

b)

\lim\limits_{x \to \infty} \left(\frac{2x + 1}{3x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(\frac{2}{3}\right)^{2x} = 0;$\underset{x\to \infty }{lim}{\left(\frac{2x+1}{3x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(\frac{2}{3}\right)}^{2x}=0;$

c)

\lim\limits_{x \to \infty} \left(\frac{3x + 1}{2x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(\frac{3}{2}\right)^{2x} = \infty;$\underset{x\to \infty }{lim}{\left(\frac{3x+1}{2x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(\frac{3}{2}\right)}^{2x}=\infty ;$

d)

\lim\limits_{x \to \infty} \left(\frac{3x + 1}{3x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(\frac{3x - 1 + 1 + 1}{3x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(1 + \frac{2}{3x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(1 + \frac{2}{3x - 1}\right)^{2x + x - x - 1 + 1} = \lim\limits_{x \to \infty} \left(1 + \frac{2}{3x - 1}\right)^{\left(3x - 1\right) \cdot \frac{2}{3} + \frac{2}{3}} = \left(e^2\right)^{\frac{2}{3}} = e^{\frac{4}{3}};$\underset{x\to \infty }{lim}{\left(\frac{3x+1}{3x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(\frac{3x-1+1+1}{3x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(1+\frac{2}{3x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(1+\frac{2}{3x-1}\right)}^{2x+x-x-1+1}=\underset{x\to \infty }{lim}{\left(1+\frac{2}{3x-1}\right)}^{\left(3x-1\right)\cdot \frac{2}{3}+\frac{2}{3}}={\left(e^2\right)}^{\frac{2}{3}}=e^{\frac{4}{3}};$

e)

\lim\limits_{x \to \infty} \left(\frac{2x + 2}{2x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(\frac{2x - 1 + 1 + 2}{2x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left(1 + \frac{3}{2x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left[\left(1 + \frac{3/2}{x}\right)^x\right]^2 = \left(e^{\frac{3}{2}}\right)^2 = e^3;$\underset{x\to \infty }{lim}{\left(\frac{2x+2}{2x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(\frac{2x-1+1+2}{2x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(1+\frac{3}{2x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left[{\left(1+\frac{3∕2}{x}\right)}^x\right]}^2={\left(e^{\frac{3}{2}}\right)}^2=e^3;$

f)

\lim\limits_{x \to \infty} \left(\frac{2x + 2}{2x - 2}\right)^{2x} = \lim\limits_{x \to \infty} \left(\frac{2x - 2 + 2 + 2}{2x - 2}\right)^{2x} = \lim\limits_{x \to \infty} \left(1 + \frac{4}{2x - 1}\right)^{2x} = \lim\limits_{x \to \infty} \left[\left(1 + \frac{4/2}{x}\right)^x\right]^2 = \left(e^{\frac{4}{2}}\right)^2 = e^4;$\underset{x\to \infty }{lim}{\left(\frac{2x+2}{2x-2}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(\frac{2x-2+2+2}{2x-2}\right)}^{2x}=\underset{x\to \infty }{lim}{\left(1+\frac{4}{2x-1}\right)}^{2x}=\underset{x\to \infty }{lim}{\left[{\left(1+\frac{4∕2}{x}\right)}^x\right]}^2={\left(e^{\frac{4}{2}}\right)}^2=e^4;$

XXX "Plusminus 1 wird bei Unendlich schon nichts ausmachen" nicht sehr elegant (Aufgaben e) und f))