Wie bestimme ich den Grenzwert ?

Wie bestimme ich den Grenzwert ?

4b.)

\(\displaystyle \lim \limits_{x\searrow 0} x^x \qquad \text{von rechts annähern}\)

mit \(\begin{array}{|rcll|} \hline x &=& 0+\dfrac{1}{n} \\ &=& \dfrac{1}{n} \\ \hline \end{array}\)

erhalten wir :  \(\displaystyle \lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} \)

Formel : \(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to \infty} f(x)\right) &=& \displaystyle\lim \limits_{x\to \infty} \ln f(x) \\ \hline \end{array}\)

Wir logarithmieren:

\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} \right) &=& \displaystyle\lim \limits_{n\to \infty} \ln \left( \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} \right)\\\\ &=& \displaystyle\lim \limits_{n\to \infty} \dfrac{1}{n}\cdot \ln \left( \dfrac{1}{n} \right)\\\\ &=& \displaystyle\lim \limits_{n\to \infty} \dfrac{ \ln \left(\dfrac{1}{n}\right) }{n} \qquad \text{Bernoulli / de l'Hospital} \\\\ &=& \displaystyle\lim \limits_{n\to \infty} \dfrac{ \Big(\ln \left(\dfrac{1}{n}\right)\Big)' }{(n)'} \quad | \quad \Big(\ln \left(\dfrac{1}{n}\right)\Big)' = \dfrac{-\dfrac{1}{n^2}}{\dfrac{1}{n}}=-\dfrac{1}{n} \qquad (n)' = 1 \\\\ &=& \displaystyle\lim \limits_{n\to \infty} \left( \dfrac{ -\dfrac{1}{n} }{1}\right) \\\\ &=& \displaystyle\lim \limits_{n\to \infty} \left( -\dfrac{1}{n} \right) \\ \hline \end{array}\)

Das Logarithmieren wird zurückgesetzt:

\(\begin{array}{|rcll|} \hline e^{ \ln \left(\displaystyle\lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} \right) } &=& e^{\displaystyle\lim \limits_{n\to \infty} \left( -\dfrac{1}{n} \right) } \\\\ \displaystyle\lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} &=& e^{\displaystyle\lim \limits_{n\to \infty} \left( -\dfrac{1}{n} \right) } \quad | \quad \displaystyle\lim \limits_{n\to \infty} \left( -\dfrac{1}{n} \right) = -0 \\\\ \displaystyle\lim \limits_{n\to \infty} \Big(\dfrac{1}{n} \Big)^{\frac{1}{n}} &=& e^{-0} \\\\ &=& e^{0} \\\\ &=& \mathbf{1} \\\\ && \boxed{ \mathbf{\displaystyle \lim \limits_{x\searrow 0} x^x = 1 } } \\ \hline \end{array}\)

Wie bestimme ich den Grenzwert ?

4c.)

\(\displaystyle \lim \limits_{x \to 0} \Big(1+\arctan(x) \Big)^{\frac{1}{x}} \)

Formel: \(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} f(x)\right) &=& \displaystyle\lim \limits_{x\to 0} \ln f(x) \\ \hline \end{array}\)
 

Wir logarithmieren:

\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} \Big(1+\arctan(x) \Big)^{\frac{1}{x}} \right) &=& \displaystyle\lim \limits_{x\to 0} \ln \left( \Big(1+\arctan(x) \Big)^{\frac{1}{x}} \right)\\\\ &=& \displaystyle\lim \limits_{x\to 0} \dfrac{ \ln \left( 1+\arctan(x) \right)}{x} \qquad \text{Bernoulli / de l'Hospital} \\\\ &=& \displaystyle\lim \limits_{x\to 0} \dfrac{ \Big( \ln \left( 1+\arctan(x) \right) \Big)'}{ (x)'} \\ && \Big( \ln \left( 1+\arctan(x) \right) \Big)' = \dfrac{\dfrac{1}{x^2+1}} {1+\arctan(x)} \qquad (x)' = 1 \\\\ &=& \displaystyle\lim \limits_{x\to 0} \dfrac{ \dfrac{\dfrac{1}{x^2+1}} {1+\arctan(x)}}{ 1} \\\\ &=& \displaystyle\lim \limits_{x\to 0} \dfrac{1} {(x^2+1)\Big(1+\arctan(x)\Big)} \\\\ &=& \dfrac{1} {(0^2+1)\Big(1+\arctan(0)\Big)} \\\\ &=& \dfrac{1} { 1\cdot(1+0)} \\\\ &=& \mathbf{1} \\ \hline \end{array} \)

Das Logarithmieren wird zurückgesetzt:

\(\begin{array}{|rcll|} \hline e^{\ln \left(\displaystyle\lim \limits_{x\to 0} \Big(1+\arctan(x) \Big)^{\frac{1}{x}} \right) } &=& e^1 \\\\ \boxed{\mathbf{ \displaystyle\lim \limits_{x\to 0} \Big(1+\arctan(x) \Big)^{\frac{1}{x}} = e}} \\ \hline \end{array}\)