What is the limit as x approaches 0 of ((cos(2x)-cos(3x)))/(x^2) ?

The limit as x approaches 0 of ((cos(2x)-cos(3x)))/(x^2) is 5/2

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limit as x approaches 0 of ((cos(2x)-cos(3x)))/(x^2)

x \to 0 lim (\frac{( cos ( 2 x ) - cos ( 3 x ) )}{x ^{2}})

\frac{5}{2}

Solution steps

x \to 0 lim (\frac{cos ( 2 x ) - cos ( 3 x )}{x ^{2}})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 2 sin ( 2 x ) + 3 sin ( 3 x )}{2 x})

= x \to 0 lim (\frac{- 2 sin ( 2 x ) + 3 sin ( 3 x )}{2 x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 4 cos ( 2 x ) + 9 cos ( 3 x )}{2})

= x \to 0 lim (\frac{- 4 cos ( 2 x ) + 9 cos ( 3 x )}{2})

Plug in the value

x = 0

= \frac{- 4 cos ( 2 \cdot 0 ) + 9 cos ( 3 \cdot 0 )}{2}

- 4 cos (2 \cdot 0) + 9 cos (3 \cdot 0) = 5

= \frac{5}{2}

\int \frac{( x + 3 )}{5 - 4 x - x ^{2}} d x

\frac{\partial}{\partial x} (sin (3 y) + 2 z e^{- 2 x})

\frac{\partial}{\partial x} (x - \frac{1}{y})

\int (2 x + 3) 2 x + 1 d x

\frac{d}{d x} (sech^{2} (3 x))

What is the limit as x approaches 0 of ((cos(2x)-cos(3x)))/(x^2) ?
The limit as x approaches 0 of ((cos(2x)-cos(3x)))/(x^2) is 5/2