Граници на функции

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[Anubis]

[tex]\fbox{3.} \quad \lim_{x \to +\infty} \frac{x^2+3x+5}{9^x} \sim \lim_{x \to +\infty} \frac{2x+3}{9^x \ln 9} \sim \lim_{x \to +\infty} \frac{2}{9^x \ln^29} = 0[/tex]

[tex]\fbox{1.} \quad \lim_{x \to 0} \frac{2-\sqrt{4 \cos 3x}}{1 - \cos 5 \sqrt{x}} = \lim_{x \to 0} \frac{2 \left ( 1 - \sqrt{\cos 3x} \right )}{2 \sin^2 \frac{5 \sqrt{x}}{2}} = \lim_{x \to 0} \frac{1 - \sqrt{\cos 3x}}{\sin^2 \frac{5 \sqrt{x}}{2}} \quad \left [ \frac{0}{0} \right ][/tex]

[tex]\left ( 1 - \sqrt{\cos 3x} \right )' = \frac{3}{2} \cdot \sin 3x \cdot \frac{1}{\sqrt{\cos 3x}}; \quad \left ( \sin^2 \frac{5 \sqrt{x}}{2} \right )' = \frac{5 \sin 5\sqrt{x}}{4\sqrt{x}}[/tex]

[tex]\Rightarrow \lim_{x \to 0} \frac{3}{2} \cdot \sin 3x \cdot \frac{1}{\sqrt{\cos 3x}} \cdot \frac{4\fbox{\sqrt{x}}}{5 \fbox{\sin 5\sqrt{x}}} \cdot \frac{\fbox{5}}{5} = \frac{6}{5} \lim_{x \to 0} \frac{\sin 3x}{\sqrt{\cos 3x}} = 0[/tex]

За другото само прилагаш формулите [tex]\left ( e^{f(x)}\right )' = e^{f(x)} \cdot f'(x)[/tex] и [tex]\left ( \ln f(x) \right )' = \frac{1}{f(x)} \cdot f'(x)[/tex].