Könnte jemand drüber schauen und sagen, ob die Grenzwerte richtig berechnet worden sind?
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(b) Berechnen Sie die Grenzwerte der Folgen:
\( \begin{array}{ll} a_{n}=\frac{\pi n^{4}-6 n^{3}+2 n^{2}-12}{\pi-n^{2}+2 \pi n^{3}-3 \pi^{2} n^{4}}, & b_{n}=\frac{\left(2 n^{2}-1\right)(n+1)}{(n-1)\left(n^{2}-4\right)} \\ c_{n}=(-1)^{n+1} \cdot \frac{(n+1)^{2}-n^{2}}{1+n^{2}}, & d_{n}=\frac{2}{n} \cdot\left(\frac{6 n^{3}-1}{n^{2}}-n\right) \end{array} \)
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b.) \( a_{n}=\frac{\pi n^{4}-6 n^{2}+2 n^{2}-12}{\pi-n^{2}+2 \pi n^{3}-3 \pi^{2} n^{4}} \)
\( \begin{array}{l} =\frac{1}{-3 \pi} \end{array} \)
\( \begin{array}{l} b_{n}=\frac{\left(2 n^{2}-1\right)(n+1)}{(n-1)\left(n^{2}-4\right)} \\ \lim \limits_{n \rightarrow \infty} \frac{\left(2 n^{2}-1\right)(n+1)}{(n-1)\left(n^{2}-4\right)}=\lim \limits_{n \rightarrow \infty} \frac{2 n^{3}+2 n^{2}-n-1}{n^{3}-n^{2}-4 n+4}=\lim \limits_{n \rightarrow \infty} \frac{n^{3}\left(2+\frac{2 \alpha^{2}}{n^{31}}-\frac{\alpha 1}{n^{22}}-\frac{1}{n^{3}}\right)}{n^{3}\left(1-\frac{10 x}{n^{2 x}}-\frac{4 x}{n^{22}}+\frac{4}{n^{3}}\right)}=\lim \limits_{n \rightarrow \infty} \frac{2+\frac{2}{-72 \rightarrow-}-\frac{1}{n^{2}}-\frac{1}{n^{2}}}{1-\frac{1}{n}-\frac{4}{n^{2}}+\frac{4}{n^{2}}}= \\ \lim \limits_{n \rightarrow \infty} \frac{2}{1}=2 \\ \rightarrow 1 \rightarrow 0 \rightarrow 0 \rightarrow 0 \\ c_{n}=(-1)^{n+1} \cdot \frac{(n+1)^{2}-n^{2}}{1+n^{2}} \\ \lim \limits_{n \rightarrow \infty}(-1)^{n+1} \cdot \frac{(n+1)^{2}-n^{2}}{1+n^{2}}=\lim \limits_{n \rightarrow \infty}(-1)^{n+1} \cdot \frac{n^{2}+2 n+1-n^{2}}{1+n^{2}}=\lim \limits_{n \rightarrow \infty}(-1)^{n+1 \cdot \frac{n\left(2+\frac{1}{n}\right)}{n^{2}\left(2\left(\frac{1}{n}+1\right)\right.}}=\lim \limits_{n \rightarrow \infty}(-1)^{n+1} \cdot \frac{1\left(2+\frac{1}{n}\right)}{\rightarrow \rightarrow 0} \rightarrow 0 \\ d_{n}=\frac{2}{n} \cdot\left(\frac{6 n^{3}-1}{n^{2}}-n\right) \\ \lim \limits_{n \rightarrow \infty} \frac{2}{n} \cdot\left(\frac{6 n^{3}-1}{n^{2}}-n\right)=\lim \limits_{n \rightarrow \infty} \frac{2}{n} \cdot\left(\frac{n^{\frac{1}{x}\left(6-\frac{1}{n^{3}}\right)}}{\frac{1}{1}}-n\right)=\lim \limits_{n \rightarrow \infty} \frac{2}{n} \cdot\left(n \cdot\left(6-\frac{1}{n^{2}}\right)-n\right)= \\ \lim \limits_{n \rightarrow \infty} \frac{2}{n} \cdot\left(6 n-\frac{n 1}{n=1}-n\right)=\lim \limits_{n \rightarrow \infty} \frac{2}{n} \cdot\left(5 n-\frac{1}{n}\right)=\lim \limits_{n \rightarrow \infty} \frac{10 x}{\infty}-\frac{2}{n^{2}}=10 \end{array} \)
