Решите уравнения: 1. lim x->0 (e^-x-e^x)/sin3x). Created by maxtaines. matematika-ru

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August 2021 1 9 Report

Решите уравнения:
1. lim x->0 (e^-x-e^x)/sin3x)

Answers & Comments

nedenispolybomy $\lim_{n \to 0} \frac{e^{-x}-e^x}{sin3x}= \lim_{n \to 0} \frac{e^{-x}(1-e^{2x})}{sin3x}= \\
 =\lim_{n \to 0} e^{-x} \lim_{n \to 0} \frac{1-e^{2x}}{sin3x}= \lim_{n \to 0} e^{-0} \lim_{n \to 0} \frac{1-e^{2x}}{sin3x}= \\
1* \lim_{n \to 0} \frac{1-e^{2x}}{sin3x}= \lim_{n \to 0} \frac{(1-e^{2x})'}{(sin3x)'}= \\
=lim_{n \to 0} \frac{-2e^{2x}}{cos3x*(3x)'}=lim_{n \to 0} \frac{-2e^{2x}}{3cos3x}= \\
= \frac{-2e^{2*0}}{3cos3*0}= \frac{-2e^0}{3*1}= \frac{-2*1}{3}= -\frac{2}{3} 
$

$y'= (\frac{cos5x}{lnx})' = \frac{(cos5x)'*lnx-cos5x*(lnx)'}{(lnx)^2}= \frac{-sin5x*(5x)'*lnx- \frac{cos5x}{x} }{ln^2x} = \\
= \frac{-sin5x*5lnx-\frac{cos5x}{x}}{ln^2x}$

$f(x)=-\frac{x^2}{x-1}$

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