Question

Limite de funcții. Calculați:

Answer 1

Răspuns:

Avem f:(0,3)→R, f(x)=$\frac{\sqrt{9-x^{2} }-3 }{x}$.

$\lim_{x \to \00} f(x)= \lim_{x \to \00} \frac{\sqrt{9-x^{2} }-3 }{x}= \lim_{x \to \00} \frac{(\sqrt{9-x^{2} }-3)(\sqrt{9-x^{2} }+3) }{x(\sqrt{9-x^{2} }+3) } = \lim_{x \to \00} \frac{9-x^{2} -9}{x(\sqrt{9-x^{2} }+3) } = \lim_{x \to \00} \frac{-x^{2} }{x(\sqrt{9-x^{2} }+3) } = \lim_{x \to \00} \frac{-x}{\sqrt{9-x^{2} }+3 } =\frac{0}{\sqrt{9-0}+3 } =0$

$\lim_{x \to \00} (1+f(x))^{\frac{1}{x} }= \lim_{x \to \00} (1+\frac{\sqrt{9-x^{2} }-3 }{x})^{\frac{1}{x} }=\lim_{x \to \00} [(1+\frac{\sqrt{9-x^{2} }-3 }{x})^{\frac{x}{\sqrt{9-x^{2} }-3} }] ^{\frac{\sqrt{9-x^{2} }-3}{x^{2} } }=e^{-\frac{1}{6} } =\frac{1}{\sqrt[6]{e} }$

$\lim_{x \to \00} \frac{\sqrt{9-x^{2} }-3}{x^{2} }=\lim_{x \to \00} \frac{(\sqrt{9-x^{2} }-3)(\sqrt{9-x^{2} }+3) }{x^{2} (\sqrt{9-x^{2} }+3) } = \lim_{x \to \00} \frac{9-x^{2} -9}{x^{2} (\sqrt{9-x^{2} }+3) } = \lim_{x \to \00} \frac{-x^{2} }{x^{2} (\sqrt{9-x^{2} }+3) } = \lim_{x \to \00} \frac{-1}{\sqrt{9-x^{2} }+3 } =\frac{-1}{\sqrt{9-0}+3 } =-\frac{1}{6}$