Question

Lt X tends to 0 cos2x-cosx/sin^2x
Solve only using L'hopital rule

Answer 1

Question

$lim_{x→1}( \frac{ \cos(2x) + \cos(x) ) }{ { \sin(x) }^{2} } )$

Answer

Using L'Hopital's rule

L'Hopital's rule : since evaluating the limits of the numerator and the denominator would result in an intermediate form

$lim_{x→0}( \frac{ \frac{dx}{dy} ( \cos(2x) - \cos(x) }{ \frac{dx}{dy} ( { \sin(x) }^{2} )} )$

Calculate the derivatives

$lim_{x→0}( \frac{ - 2 \sin(2x) - \sin(x) }{ \sin(2x) } )$

Expand the expression

$lim_{x→0}( \frac{ - 2 \times 2 \sin(x) \cos(x) + \sin(x) }{2 \sin(x) \cos(x) } )$

Factor the expression

$lim_{x→0}( \frac{ - \sin(x)(2 \times 2 \cos(x) - 1)}{2 \sin(x) \cos(x) } )$

Reduce multiply

$lim_{x→0}( \frac{ - (4 \cos(x) - 1) }{2 \cos(x) } )$

Rewrite the fraction

$lim_{x→0}( \frac{1 - 4 \cos(x) }{2 \cos(x) } )$

Evaluate the limit & simplify the expression

$lim_{x→0}( \frac{1 - 4 \cos(0) }{2 \cos(0) } )$

$= \frac{ - 3}{2}$