What is the limit as x approaches 0 of (cos(3x)-cos(4x))/(x^2) ?

The limit as x approaches 0 of (cos(3x)-cos(4x))/(x^2) is 7/2

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limit as x approaches 0 of (cos(3x)-cos(4x))/(x^2)

x \to 0 lim (\frac{cos ( 3 x ) - cos ( 4 x )}{x ^{2}})

\frac{7}{2}

Decimal

3.5

Solution steps

x \to 0 lim (\frac{cos ( 3 x ) - cos ( 4 x )}{x ^{2}})

Apply L'Hopital's Rule

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

Apply L'Hopital's Rule

= x \to 0 lim (\frac{- 9 cos ( 3 x ) + 16 cos ( 4 x )}{2})

Plug in the value

x = 0

= \frac{- 9 cos ( 3 \cdot 0 ) + 16 cos ( 4 \cdot 0 )}{2}

- 9 cos (3 \cdot 0) + 16 cos (4 \cdot 0) = 7

= \frac{7}{2}

What is the limit as x approaches 0 of (cos(3x)-cos(4x))/(x^2) ?
The limit as x approaches 0 of (cos(3x)-cos(4x))/(x^2) is 7/2