Odpowiedź :
[tex]\LARGE\begin{aligned}&\lim_{n\to\infty}\left(1+\dfrac{1}{n-5}\right)^n=\lim_{n\to\infty}\left(\left(1+\dfrac{1}{n-5}\right)^{n-5}\right)^{\frac{n}{n-5}}=\\&=e^{\displaystyle \lim_{n\to\infty}\frac{n}{n-5}}=e^{\displaystyle \lim_{n\to\infty}\frac{1}{1-\frac{5}{n}}}=e^1=e\end[/tex]
[tex]\lim_{n \to \infty} (1+\frac{1}{n-5})^n= \lim_{n \to \infty} (1+\frac{1}{n-5})^{n-5} \cdot(1+\frac{1}{n-5})^5=\\ \\ \stackrel{m:=n-5}{=} \lim_{m \to \infty} (1+\frac{1}{m})^{m} \cdot (1+\frac{1}{m})^5=e \cdot 1=e[/tex]