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

The limit as x approaches 0 of (cos(ax)-cos(bx))/(x^2) is (-a^2+b^2)/2

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

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

\frac{- a ^{2} + b ^{2}}{2}

Solution steps

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- a ^{2} cos ( a x ) + b ^{2} cos ( b x )}{2})

= x \to 0 lim (\frac{- a ^{2} cos ( a x ) + b ^{2} cos ( b x )}{2})

Plug in the value

x = 0

= \frac{- a ^{2} cos ( a \cdot 0 ) + b ^{2} cos ( b \cdot 0 )}{2}

- a^{2} cos (a \cdot 0) + b^{2} cos (b \cdot 0) = - a^{2} + b^{2}

= \frac{- a ^{2} + b ^{2}}{2}

f (x) = 5 x^{e}

y^{^{'}} = 1 - sin (y), y (0) = 0

\int_{- 1}^{1} 3 x^{2} (2 x^{3} + 1)^{2} d x

\int x^{3} - 3 x^{2} d x

x \to 2 + lim (\frac{2}{( x - 2 ) ^{5}})

What is the limit as x approaches 0 of (cos(ax)-cos(bx))/(x^2) ?
The limit as x approaches 0 of (cos(ax)-cos(bx))/(x^2) is (-a^2+b^2)/2