Granice ciągu

[JJ35]

\(\lim_{x\to \infty }(1+ \frac{3}{x})^x\)
\(\lim_{x\to \infty }(\frac{x+1}{x-3})^x\)
\(\lim_{x\to \infty }( \frac{x^2+3}{x^2+7} )^x^2\)

[Lbubsazob]

W każdej z tych granic mamy symbol nieoznaczony typu \(1^\infty\), więc trzeba skorzystać z twierdzenia o liczbie \(e\).

Zad. 1
\(\lim_{x\to \infty } \left( 1+ \frac{3}{x} \right) ^x=\lim_{x\to \infty } \left[\left( 1+ \frac{3}{x} \right) ^x \right]^{ \frac{x}{3} \cdot \frac{3}{x}}=e^3\)
bo \(\lim_{x\to\infty} \left( 1+ \frac{3}{x} \right) ^{ \frac{x}{3}}=e\)

[Lbubsazob]

Zad. 2
\(\lim_{x\to \infty } \left( \frac{x+1}{x-3} \right)^x=\lim_{x\to \infty } \left( \frac{x-3+4}{x-3} \right)^x=\lim_{x\to \infty } \left(1+ \frac{4}{x-3} \right)^x=\lim_{x\to \infty } \left[\left(1+ \frac{4}{x-3} \right)^x \right]^{ \frac{4}{x-3} \cdot \frac{x-3}{4}} = \\ = e^{\lim_{x\to \infty } \frac{4x}{x-3}} =e^4\)

[Lbubsazob]

Zad. 3
\(\lim_{x\to\infty} \left( \frac{x^2+3}{x^2+7} \right)^{x^2} = \lim_{x\to\infty} \left( \frac{x^2+7-4}{x^2+7} \right)^{x^2} =\lim_{x\to\infty} \left( 1+\frac{-4}{x^2+7} \right)^{x^2} =\lim_{x\to\infty} \left[\left( 1+\frac{-4}{x^2+7} \right)^{x^2}\right] ^{ \frac{x^2+7}{-4} \cdot \frac{-4}{x^2+7}} = \\ =e^{ \lim_{x\to \infty} \frac{-4x^2}{x^2+7}}=e^{-4}\)