What is the limit as x approaches 0 of ((cos(4x)-cos(6x)))/(cos(2x)sin^2(5x)) ?

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limit as x approaches 0 of ((cos(4x)-cos(6x)))/(cos(2x)sin^2(5x))

x \to 0 lim (\frac{( cos ( 4 x ) - cos ( 6 x ) )}{cos ( 2 x ) sin ^{2} ( 5 x )})

\frac{2}{5}

Decimal Notation

0.4

Solution steps

x \to 0 lim (\frac{cos ( 4 x ) - cos ( 6 x )}{cos ( 2 x ) sin ^{2} ( 5 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 4 sin ( 4 x ) + 6 sin ( 6 x )}{- 2 sin ^{2} ( 5 x ) sin ( 2 x ) + 5 sin ( 10 x ) cos ( 2 x )})

= x \to 0 lim (\frac{- 4 sin ( 4 x ) + 6 sin ( 6 x )}{- 2 sin ^{2} ( 5 x ) sin ( 2 x ) + 5 sin ( 10 x ) cos ( 2 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 16 cos ( 4 x ) + 36 cos ( 6 x )}{- 20 sin ( 10 x ) sin ( 2 x ) - 4 sin ^{2} ( 5 x ) cos ( 2 x ) + 50 cos ( 10 x ) cos ( 2 x )})

= x \to 0 lim (\frac{- 16 cos ( 4 x ) + 36 cos ( 6 x )}{- 20 sin ( 10 x ) sin ( 2 x ) - 4 sin ^{2} ( 5 x ) cos ( 2 x ) + 50 cos ( 10 x ) cos ( 2 x )})

Plug in the value

x = 0

= \frac{- 16 cos ( 4 \cdot 0 ) + 36 cos ( 6 \cdot 0 )}{- 20 sin ( 10 \cdot 0 ) sin ( 2 \cdot 0 ) - 4 sin ^{2} ( 5 \cdot 0 ) cos ( 2 \cdot 0 ) + 50 cos ( 10 \cdot 0 ) cos ( 2 \cdot 0 )}

Simplify

\frac{- 16 cos ( 4 \cdot 0 ) + 36 cos ( 6 \cdot 0 )}{- 20 sin ( 10 \cdot 0 ) sin ( 2 \cdot 0 ) - 4 sin ^{2} ( 5 \cdot 0 ) cos ( 2 \cdot 0 ) + 50 cos ( 10 \cdot 0 ) cos ( 2 \cdot 0 )} : \frac{2}{5}

= \frac{2}{5}

y^{^{'}} + y \cdot sin (x) = sin (x)

e^{- x} x - 2 e^{- x}

(cos (x^{2}))^{^{'}}

\int cos (a)

\int x^{2} + 10 d x

What is the limit as x approaches 0 of ((cos(4x)-cos(6x)))/(cos(2x)sin^2(5x)) ?
The limit as x approaches 0 of ((cos(4x)-cos(6x)))/(cos(2x)sin^2(5x)) is 2/5