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Makovka662 [10]
2 years ago
14

What is the problem of this solving?!

Mathematics
1 answer:
nika2105 [10]2 years ago
3 0
     This question can be solved primarily by L'Hospital Rule and the Product Rule.

y= \lim_{x \to 0}  \frac{x^2cos(x)-sin^2(x)}{x^4}
 
     I) Product Rule and L'Hospital Rule:

y= \lim_{x \to 0} \frac{[2xcos(x)-x^2sin(x)]-2sin(x)cos(x)}{4x^3}
 
     II) Product Rule and L'Hospital Rule:

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}
 
     III) Product Rule and L'Hospital Rule:

]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}
 
     IV) Product Rule and L'Hospital Rule:

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}
 
     V) Using the Definition of Limit:

y= \frac{-6*1-4*1-2*1+8*1^2}{24}  \\ y= \frac{-4}{24}  \\ \boxed {y= \frac{-1}{6} }
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