Question

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

Answer 1

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

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

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