c

$$\underset{n\to \infty }{\lim }{n}^2-\sqrt{n^4+3n^2+2}=\text{?}$$

Optionen

• Divergenz beweisen $\to$ (s7)
• Grenzwert berechnen
  • Sandwichtheorem
  • Auf wichtige Reihen zurückführen $\to$ (s15)
  • Auf wichtige Grenzwerte zurückführen $\to$ (s8)
  • $\lim$-Regeln $\to$ (s7)

Bearbeitung Umformen mit 3. binom. Formel

$$\begin{array}{rl}\underset{n\to \infty }{\lim }{n}^2-\sqrt{n^4+3n^2+2}& =\underset{n\to \infty }{\lim }{n}^2-\sqrt{n^4+3n^2+2}\cdot \frac{n^2+\sqrt{n^4+3n^2+2}}{n^2+\sqrt{n^4+3n^2+2}}\\ & =\underset{n\to \infty }{\lim }\frac{(n^2-\sqrt{n^4+3n^2+2})\cdot (n^2+\sqrt{n^4+3n^2+2})}{n^2+\sqrt{n^4+3n^2+2}}\\ & \stackrel{(4)}{=}\underset{n\to \infty }{\lim }\frac{(n^2{)}^2-(\sqrt{n^4+3n^2+2}{)}^2}{n^2+\sqrt{n^4+3n^2+2}}\\ & =\underset{n\to \infty }{\lim }\frac{n^4-(n^4+3n^2+2)}{n^2+\sqrt{n^4+3n^2+2}}\\ & =\underset{n\to \infty }{\lim }\frac{-3n^2-2}{n^2+\sqrt{n^4+3n^2+2}}\\ & =\underset{n\to \infty }{\lim }\frac{-3n^2-2}{n^2+\sqrt{n^4+3n^2+2}}\cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}}\\ & =\underset{n\to \infty }{\lim }\frac{\frac{-3n^2-2}{n^2}}{\frac{n^2+\sqrt{n^4+3n^2+2}}{n^2}}\\ & =\underset{n\to \infty }{\lim }\frac{\frac{-3n^2}{n^2}-\frac{2}{n^2}}{\frac{n^2}{n^2}+\frac{\sqrt{n^4+3n^2+2}}{n^2}}\\ & =\underset{n\to \infty }{\lim }\frac{\frac{-3n^2}{n^2}-\frac{2}{n^2}}{\frac{n^2}{n^2}+\sqrt{\frac{n^4+3n^2+2}{(n^2{)}^2}}}\\ & =\underset{n\to \infty }{\lim }\frac{-3-\frac{2}{n^2}}{1+\sqrt{\frac{n^4}{n^4}+\frac{3n^2}{n^4}+\frac{2}{n^4}}}\\ & =\underset{n\to \infty }{\lim }\frac{-3-\frac{2}{n^2}}{1+\sqrt{1+\frac{3}{n^2}+\frac{2}{n^4}}}\\ & \stackrel{(2)}{=}\frac{{\lim }_{n\to \infty }-3-\frac{2}{n^2}}{{\lim }_{n\to \infty }1+\sqrt{1+\frac{3}{n^2}+\frac{2}{n^4}}}\\ & =\frac{-3}{2}\end{array}$$