What is the limit as x approaches 0 of (cos(4x)-cos(2x))/(x^2) ?

The limit as x approaches 0 of (cos(4x)-cos(2x))/(x^2) is -6

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limit as x approaches 0 of (cos(4x)-cos(2x))/(x^2)

x \to 0 lim (\frac{cos ( 4 x ) - cos ( 2 x )}{x ^{2}})

- 6

Solution steps

x \to 0 lim (\frac{cos ( 4 x ) - cos ( 2 x )}{x ^{2}})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 4 sin ( 4 x ) + 2 sin ( 2 x )}{2 x})

= x \to 0 lim (\frac{- 4 sin ( 4 x ) + 2 sin ( 2 x )}{2 x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 16 cos ( 4 x ) + 4 cos ( 2 x )}{2})

= x \to 0 lim (\frac{- 16 cos ( 4 x ) + 4 cos ( 2 x )}{2})

Plug in the value

x = 0

= \frac{- 16 cos ( 4 \cdot 0 ) + 4 cos ( 2 \cdot 0 )}{2}

Simplify

\frac{- 16 cos ( 4 \cdot 0 ) + 4 cos ( 2 \cdot 0 )}{2} : - 6

= - 6

\frac{d}{d x} (\frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}})

\int (\frac{11}{x} + \frac{x}{11}) d x

y^{^{''}} - 14 y^{^{'}} + 53 y = 0

\int 20 cos (10 t + \frac{π}{6}) + 2 d t

\int \frac{cos ( 3 x )}{sin ( 3 x )} d x

What is the limit as x approaches 0 of (cos(4x)-cos(2x))/(x^2) ?
The limit as x approaches 0 of (cos(4x)-cos(2x))/(x^2) is -6