granica z e

[waliant]

Wiemy, że \(\displaystyle{ x _{n} \rightarrow \infty \Rightarrow \lim_{n \to \infty } \left( 1+ \frac{1}{x _{n} } \right) ^{x ^{n} } =e}\). Jak pokazać, że zachodzi:
\(\displaystyle{ x _{n} \rightarrow \infty \Rightarrow \lim_{n \to \infty } \left( 1+ \frac{a}{x _{n} } \right) ^{x ^{n} }=e ^{a}}\)?

[sneik555]

\(\displaystyle{ \lim_{n \to \infty } \left( 1+ \frac{a}{x _{n} } \right) ^{x ^{n} }=\lim_{n \to \infty } \left( 1+ \frac{1}{\frac{x _{n}}{a} } \right) ^{\frac{x ^{n}}{a}a }=e^a}\)