Няколко задачи от граници

[Black_Devil]

[tex]\lim_{x\to\0} \frac{e^{x} - e^{-x} - 2x}{ x - sinx }[/tex]

[tex]\lim_{x\to\0} \frac{ln(1 - cosx)}{ln x}[/tex]

[tex]\lim_{x\to\ \infty } (\sqrt{x^{2} + x} - x)[/tex]

[tex]\lim_{x\to\0} x^{2} e^{ \frac{1}{x^{2}}}[/tex]

[tex]\lim_{x\to\0}(x + 2^{x})^{\frac {1}{x}}[/tex]

[Anubis]

[tex]1. \quad \lim_{x \to 0} \frac{e^x-e^{-x}-2x}{x - \sin x} \, \left [ \frac{0}{0} \right ] = \lim_{x \to 0} \frac{e^x+e^{-x}-2}{1 - \cos x} \, \left [ \frac{0}{0} \right ] = \lim_{x \to 0} \frac{e^x-e^{-x}}{\sin x} \left [ \frac{0}{0} \right ] = \lim_{x \to 0} \frac{e^x+e^{-x}}{\cos x} = 2[/tex]

[tex]2. \quad \lim_{x \to 0} \frac{\ln (1 - \cos x)}{\ln x} \quad : \quad \lim_{x \to 0} \ln(1 - \cos x) = \ln 0 = -\infty; \quad \lim_{x \to 0} \ln x = \ln 0 = -\infty[/tex]

Получава се неопределеност от вида [tex]\left [ \frac{\infty}{\infty} \right ][/tex].

[tex]\lim_{x \to 0} \frac{\ln (1 - \cos x)}{\ln x} = \lim_{x \to 0} \frac{1}{1 - \cos x} \cdot \sin x \cdot x = \lim_{x \to 0} \frac{x \sin x}{1 - \cos x}[/tex]

[tex]1 - \cos x = 1 - \left ( \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} \right ) = 2 \sin^2 \frac{x}{2} \Rightarrow \lim_{x \to 0} \frac{\cancel{2}x \cancel{\sin\frac{x}{2}} \cos \frac{x}{2}}{\cancel{2} \cancel{\sin \frac{x}{2}} \sin \frac{x}{2}} = \lim_{x \to 0} \cos \frac{x}{2} \cdot \lim_{x \to 0} \frac{x}{\sin \frac{x}{2}} = \lim_{x \to 0} 2\frac{\frac{x}{2}}{\sin \frac{x}{2}} = 2[/tex]

[tex]3. \quad \lim_{x \to +\infty} \left ( \sqrt{x^2+x} - x \right ) = \lim_{x \to +\infty} \frac{x^2+x-x^2}{\sqrt{x^2+x} + x} = \lim_{x \to +\infty} \frac{x}{x\sqrt{1+\frac{1}{x}}+x} = \lim_{x \to +\infty} \frac{1}{\sqrt{1+\frac{1}{x}}+1} = \frac{1}{2}[/tex]

[tex]4. \quad \lim_{x \to 0}x^2 e^{\frac{1}{x^2}} = \lim_{x \to 0} \frac{e^{\frac{1}{x^2}}}{x^{-2}} = \lim_{x \to 0} \frac{e^{\frac{1}{x^2}}}{\frac{1}{x^2}} \left [ \frac{\infty}{\infty} \right ] = \lim_{x \to 0} \frac{ \left ( \frac{1}{x^2} \right )' e^{\frac{1}{x^2}}}{\left ( \frac{1}{x^2} \right )'} = \lim_{x \to 0} e^{\frac{1}{x^2}} = +\infty[/tex]

[tex]5. \quad \lim_{x \to 0} \left ( x+2^x \right )^{\frac{1}{x}} \quad \left [ 1^{\infty} \right ][/tex]

Когато при пресмятането на [tex]\lim_{x \to x_{0}}\left ( f(x) \right )^{g(x)}[/tex] след граничен преход се получи

неопределеност [tex]\left [ 1^{\infty} \right ][/tex], границата е равна на [tex]e^a, \quad a = \lim_{x \to x_{0}} g(x) \left [ f(x) - 1 \right ][/tex]. Тук е така.

[tex]f(x) = x+2^x, \, g(x) = \frac{1}{x} \Rightarrow \lim_{x \to 0} \frac{1}{x} \left ( x+2^x-1 \right ) = \lim_{x \to 0} \frac{x+2^x-1}{x} \left [ \frac{0}{0} \right ] = \lim_{x \to 0} (x+2^x)' = \lim_{x \to 0} (1+2^x \ln 2) = 1 + \ln 2[/tex]

[tex]a = 1 + \ln 2 \Rightarrow e^a = e^{1 + \ln 2} = e \cdot e^{\ln 2} = 2e[/tex]

Спомени от ДИС 1...

[Black_Devil]

Леле поздравления пич аз на това само кодовете му не мога да напиша ... много ти благодаря!!!