Question

What is the problem of this solving?!

Answer 1

     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:

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