Bestimmen Sie Analysis

Question

Bestimmen Sie
A
\( \lim \limits_{n \rightarrow \infty} \frac{(n+3)^{3}}{(3 n+1)^{3}} \)


B
\( \lim \limits_{n \rightarrow \infty} \frac{\sqrt{2^{n}+4^{n}}}{2^{n-1}} \)

Answer 1

Aloha :)

$$\frac{(n+3)^3}{(3n+1)^3}=\left(\frac{n+3}{3n+1}\right)^3=\left(\frac{n+\frac{1}{3}+\frac{8}{3}}{3n+1}\right)^3=\left(\frac{n+\frac{1}{3}}{3n+1}+\frac{\frac{8}{3}}{3n+1}\right)^3$$$$\phantom{\frac{(n+3)^3}{(3n+1)^3}}=\left(\frac{\frac{1}{3}(3n+1)}{3n+1}+\frac{8}{3(3n+1)}\right)^3=\left(\frac{1}{3}+\frac{8}{3(3n+1)}\right)^3\to\left(\frac{1}{3}\right)^3=\frac{1}{27}$$

$$\frac{\sqrt{2^n+4^n}}{2^{n-1}}=\frac{\sqrt{2^n+(2^2)^n}}{2^{n-1}}=\frac{\sqrt{2^n+2^{2n}}}{2^{n-1}}=\frac{\sqrt{2^{2n}\left(\frac{1}{2^n}+1\right)}}{2^{n-1}}=\frac{\sqrt{2^{2n}}\sqrt{\frac{1}{2^n}+1}}{2^{n-1}}$$$$\phantom{\frac{\sqrt{2^n+4^n}}{2^{n-1}}}=\frac{\sqrt{(2^n)^2}\sqrt{\frac{1}{2^n}+1}}{2^{n-1}}=\frac{2^n\sqrt{\frac{1}{2^n}+1}}{2^{n-1}}=2\sqrt{1+\frac{1}{2^n}}\to2$$