![y= \lim_{x \to 0} \frac{[2xcos(x)-x^2sin(x)]-2sin(x)cos(x)}{4x^3}](/tpl/images/0184/2974/d1dcf.png)
![y= \lim_{x \to 0} \frac{[-2xsin(x)+2cos(x)]-[2xsin(x)+x^2cos(x)]-[2cos^2(x)-2sin^2(x)]}{12x^2} \\ y= \lim_{x \to 0} \frac{2cos(x)-4xsin(x)-x^2cos(x)-2cos^2(x)+2sin^2(x)}{12x^2}](/tpl/images/0184/2974/79375.png)
![]y= \alpha + \beta \\ \\ \alpha =\lim_{x \to 0} \frac{-2sin(x)-[4sin(x)+4xcos(x)]-[2xcos(x)-x^2sin(x)]}{24x} \\ \beta = \lim_{x \to 0} \frac{4sin(x)cos(x)+4sin(x)cos(x)}{24x} \\ \\ y = \lim_{x \to 0} \frac{-6sin(x)-4xcos(x)-2xcos(x)+x^2sin(x)+8sin(x)cos(x)}{24x}](/tpl/images/0184/2974/6bb12.png)
![y = \phi + \varphi \\ \\ \phi = \lim_{x \to 0} \frac{-6cos(x)-[-4xsin(x)+4cos(x)]-[2cos(x)-2xsin(x)]}{24x} \\ \varphi = \lim_{x \to 0} \frac{[2xsin(x)+x^2cos(x)]+[8cos^2(x)-8sin(x)]}{24x}](/tpl/images/0184/2974/0b92d.png)
What is the problem of this solving? !
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Mathematics, 21.08.2019 08:30 mathman783
What is the problem of this solving? !
Answers: 1
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