Programemos Cosas: Matemática II

33.
$\\[0\parindent] \lim_{x \to 0} \frac {\sin^2 \frac{x}{3}}{x^2} = \\[2\baselineskip] = \lim_{x \to 0} \frac {\sin \frac{x}{3}}{x} . \lim_{x \to 0} \frac {\sin \frac{x}{3}}{x} = \\[2\baselineskip] = \frac{1}{3}.\frac{1}{3} = \frac{1}{9}$

34.
$\\[0\parindent] \lim_{x \to +0} \frac {x}{\sqrt{1-\cos x}} = \\[2\baselineskip] = = \lim_{x \to +0} \frac {x \sqrt{1+\cos x}}{\sqrt{1-\cos x} \sqrt{1+\cos x}} = \\[2\baselineskip] = \lim_{x \to +0} \frac {x \sqrt{1+\cos x}}{\sqrt{1-\cos x} \sqrt{1+\cos x}} = \\[2\baselineskip] = \lim_{x \to +0} \frac {x \sqrt{1+\cos x}}{\sqrt{1-\cos^2 x}} = \\[2\baselineskip] = \lim_{x \to +0} \frac {x \sqrt{1+\cos x}}{\sqrt{\sin^2 x + \cos^2 x - \cos^2 x}} = \\[2\baselineskip] = \lim_{x \to +0} \frac {x \sqrt{1+\cos x}}{\sqrt{\sin^2 x}} = \\[2\baselineskip] = \lim_{x \to +0} \frac {x \sqrt{1+\cos x}}{\sin x} = \\[2\baselineskip] = \lim_{x \to +0} \frac {x}{\sin x} . \lim_{x \to +0} x \sqrt{1+\cos x} = \\[2\baselineskip] = \lim_{x \to +0} \frac {1}{\frac{\sin x}{x}} . \lim_{x \to +0} x \sqrt{1+\cos x} = \sqrt{2}$

35.
$\\[0\parindent] \lim_{x \to 0} x . \cot x = \\[2\baselineskip] = \lim_{x \to 0} x . \frac {\cos x}{\sin{x}} = \\[2\baselineskip] = \lim_{x \to 0} \frac {1}{\frac{\sin x}{x}} . \cos x = 1$

36.
$\\[0\parindent] \lim_{v \to \frac{\pi}{3}} \frac {1-2\cos v}{\sin(v - \frac{\pi}{3})} = \\[2\baselineskip] = \lim_{w \to 0} \frac {1-2\cos (w+\frac{\pi}{3})}{\sin w} = \\[2\baselineskip] = \lim_{w \to 0} \frac {1-2\cos w \cos \frac{\pi}{3}-2\sin w \sin \frac{\pi}{3}}{\sin w} = \\[2\baselineskip] = \lim_{w \to 0} \frac {1-\cos w-2\frac{\sqrt{3}}{2}\sin w }{\sin w} = \\[2\baselineskip] = \lim_{w \to 0} \frac {1-\cos w}{\sin w} - 2 \frac {\sqrt{3}\sin w}{2 \sin w} = \\[2\baselineskip] = \lim_{w \to 0} \frac {(1-\cos w)(1+\cos w)}{\sin w (1+\cos w)} - \sqrt{3} = \\[2\baselineskip] = \lim_{w \to 0} \frac {1-\cos^2w}{\sin w+\sin w\cos w} - \sqrt{3} = \\[2\baselineskip] = \lim_{w \to 0} \frac {1-\cos^2 w}{\sin w+\sin w\cos w} - \sqrt{3} = \\[2\baselineskip] = \lim_{w \to 0} \frac {\sin^2 w}{\sin w+\sin w\cos w} - \sqrt{3} = \\[2\baselineskip] = \lim_{w \to 0} \frac {\sin w}{1+\cos w} - \sqrt{3} = -\sqrt{3}, w=v-\frac{\pi}{3}, w \to 0 \Leftrightarrow v \to \frac{\pi}{3}$

37.
$\\[0\parindent] \lim_{z \to 1} (1-z) \tan \frac{z \pi}{2} = \\[2\baselineskip] = \lim_{\theta \to \pi} (1-\frac{\theta}{\pi}) \tan \frac{\theta}{2}= \\[2\baselineskip] = \lim_{\theta \to \pi} (1-\frac{\theta}{\pi}) \frac{1-\cos \theta}{\sin \theta} = \\[2\baselineskip] = \frac{1}{\pi} \lim_{\theta \to \pi, \alpha \to 0} \frac{\alpha}{\sin \alpha} (1-\cos \theta) = \frac{2}{\pi}\\[2\baselineskip]\theta=\pi z\\[2\baselineskip]\alpha=\pi-\theta$

38.
$\\[0\parindent] \lim_{x \to 0} \frac{2 \arcsin x}{3x} = \\[2\baselineskip] = \frac{2}{3} \lim_{y \to 0} \frac{y}{\sin y} = \frac{2}{3} \\[2\baselineskip] \\[2\baselineskip]y=\arcsin x, \sin y = x$

39.
$\\[0\parindent] \lim_{x \to 0} \frac{\sin(\alpha+x)-\sin(\alpha-x)}{x} = \\[2\baselineskip] = \lim_{x \to 0} \frac{\sin \alpha \cos x + \cos \alpha \sin x - \sin \alpha \cos x + \cos \alpha \sin x }{x} = \\[2\baselineskip] = \lim_{x \to 0} \frac{2 \cos \alpha \sin x }{x} = 2 \cos \alpha$

40.
$\\[0\parindent] \lim_{x \to 0} \frac{\tan x-\sin x}{x^3} = \\[2\baselineskip] = \lim_{x \to 0} \frac{\tan x-\sin x}{x^3} = \\[2\baselineskip] = \lim_{x \to 0} \frac{\sin x}{x} \frac{\frac{1}{\cos x}-1}{x^2}= \\[2\baselineskip] = \lim_{x \to 0} \frac{\sin x}{x} \frac{\frac{1-\cos x}{\cos x}}{x^2}= \\[2\baselineskip] = \lim_{x \to 0} \frac{\sin x}{x} \frac{\frac{(1-\cos x)(1+\cos x)}{\cos x (1+\cos x)}}{x^2}= \\[2\baselineskip] = \lim_{x \to 0} \frac{\sin x}{x} \frac{\frac{1-\cos^2 x}{\cos x+\cos^2 x}}{x^2}= \\[2\baselineskip] = \lim_{x \to 0} \frac{\sin x}{x} \frac{\frac{\sin^2 x}{\cos x+\cos^2 x}}{x^2}= \\[2\baselineskip] = \lim_{x \to 0} \frac{\sin x}{x} \frac{1}{\cos x+\cos^2 x} \frac{\sin^2 x}{x^2} = \frac{1}{2}$

41.
$\\[0\parindent] \lim_{x \to \infty} (1+\frac{2}{x})^x = \\[2\baselineskip] = \lim_{y \to \infty} (1+\frac{1}{y})^{2y} = \\[2\baselineskip] = \lim_{y \to \infty} (1+\frac{1}{y})^y \lim_{y \to \infty} (1+\frac{1}{y})^y = e^2 \\[2\baselineskip] \\[2\baselineskip] 2y=x \\[2\baselineskip] x \to \infty \Leftrightarrow y \to \infty$

42.
$\\[0\parindent] \lim_{x \to \infty} (1-\frac{1}{x})^x = \\[2\baselineskip] = \lim_{y \to \infty} (1+\frac{1}{y})^{-y} = \\[2\baselineskip] = \lim_{y \to \infty} \frac{1}{(1+\frac{1}{y})^{y}} = \\[2\baselineskip] = \frac{\lim_{y \to \infty} 1}{\lim_{y \to \infty} (1+\frac{1}{y})^{y}} = \frac{1}{e} \\[2\baselineskip] \\[2\baselineskip] y=-x \\[2\baselineskip] y \to \pm \infty \Leftrightarrow x \to \mp \infty$

43.
$\\[0\parindent] \lim_{x \to \infty} (\frac{x}{1+x})^x = \\[2\baselineskip] = \lim_{x \to \infty} (\frac{1+x}{x})^{-x} = \\[2\baselineskip] = \lim_{x \to \infty} (\frac{1}{x}+1)^{-x} = \frac{1}{e}$
Ver 42

44.
$\\[0\parindent] \lim_{n \to \infty} (1+\frac{1}{n})^{n+5} = \\[2\baselineskip] = \lim_{n \to \infty} (1+\frac{1}{n})^n \lim_{n \to \infty} (1+\frac{1}{n})^5 = e$

45.
$\\[0\parindent] \lim_{n \to \infty} n [\ln (n+1) - \ln n] = \\[2\baselineskip] = \lim_{n \to \infty} n \ln \frac{(n+1)}{n} = \\[2\baselineskip] = \lim_{n \to \infty} \ln (1+\frac{1}{n})^n = 1$

46.
$\\[0\parindent] \lim_{x \to \frac{\pi}{2}} (1 + \cos x)^{3 \sec x} = \\[2\baselineskip] = \lim_{x \to \frac{\pi}{2}} (1 + \frac{1}{\sec x})^{3 \sec x} = \\[2\baselineskip] = \lim_{y \to \infty} (1 + \frac{1}{y})^{3y} = e^3 \\[2\baselineskip] \\[2\baselineskip] y=\sec x, \lim_{x \to \frac{\pi}{2}} y = \infty$

47.
$\\[0\parindent] \lim_{x \to 0} \frac{\ln (1+\alpha x)}{x} = \\[2\baselineskip] = \lim_{x \to 0} \ln (1+\alpha x)^\frac{1}{x} = \\[2\baselineskip] = \lim_{y \to \infty} \ln (1+\frac{1}{y})^{\alpha y} = \alpha \\[2\baselineskip] \frac{1}{y} = \alpha x, \lim_{x \to 0} y = \infty$

48.
$\\[0\parindent] \lim_{x \to \infty} (\frac{2x+3}{2x+1})^{x+1} = \\[2\baselineskip] = \lim_{y \to \infty} (\frac{y-1+3}{y})^{\frac{y}{2}+\frac{1}{2}} = \\[2\baselineskip] = \lim_{y \to \infty} (1+\frac{2}{y})^{\frac{y}{2}+\frac{1}{2}} = \\[2\baselineskip] = \lim_{z \to \infty} (1+\frac{1}{z})^{z+\frac{1}{2}} = \\[2\baselineskip] = \lim_{z \to \infty} (1+\frac{1}{z})^z (1+\frac{1}{z})^\frac{1}{2} = e \\[2\baselineskip] y=2x+1, z=\frac{y}{2}$

49.
$\\[0\parindent] \lim_{x \to \0} (1+3\tan^2 x)^{\cot^2 x} = \\[2\baselineskip] = \lim_{y \to \infty} (1+\frac{1}{y})^{3y} = e^3 \\[2\baselineskip] \frac{1}{y}=3\tan^2 x, \lim_{x \to 0} y = \infty$

50.
$\\[0\parindent] \lim_{m \to \infty} (\cos \frac{x}{m})^m = \\[2\baselineskip] = \lim_{m \to \infty} (\sqrt{1-\sin^2\frac{x}{m}})^m = \\[2\baselineskip] = \lim_{m \to \infty} \sqrt{(1-\sin^2\frac{x}{m})^m} \\[2\baselineskip] (*1) \lim_{m \to \infty} \sqrt{(1-\frac{x^2}{m^2})^m} \le \lim_{m \to \infty} \sqrt{(1-\sin^2\frac{x}{m})^m} \le \lim_{m \to \infty} \sqrt{1^m}=1 \\[2\baselineskip] \lim_{m \to \infty}\sqrt{(1-\frac{x^2}{m^2})^m} (1*) = \\[2\baselineskip] = \lim_{m \to \infty} \sqrt{(1+\frac{x}{m})^m (1-\frac{x}{m})^m} = \\[2\baselineskip] = \lim_{m \to \infty} \sqrt{e^x e^{-x}} = 1 \\[2\baselineskip] \therefore \lim_{m \to \infty} \sqrt{(1-\sin^2\frac{x}{m})^m}=1$

Bookboon
Wikibooks, subjects.
Differential and Integral Calculus : N. Piskunov : Free Download & Streaming : Internet Archive:
What Is an Algorithm?
Algorithms - Wikibooks, open books for an open world
Computers and Information Processing - PHI Learning
Introduction to algorithms 3rd edition
Solutions for Introduction to algorithms second edition PDF - Ebookily.org

Programemos Cosas

miércoles, 8 de enero de 2014

Matemática II

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