What is the limit as x approaches 0 of (1-cos(8x))/(4x^2) ?

The limit as x approaches 0 of (1-cos(8x))/(4x^2) is 8

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limit as x approaches 0 of (1-cos(8x))/(4x^2)

x \to 0 lim (\frac{1 - cos ( 8 x )}{4 x ^{2}})

8

Solution steps

x \to 0 lim (\frac{1 - cos ( 8 x )}{4 x ^{2}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{4} \cdot x \to 0 lim (\frac{1 - cos ( 8 x )}{x ^{2}})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{8 sin ( 8 x )}{2 x})

= \frac{1}{4} \cdot x \to 0 lim (\frac{8 sin ( 8 x )}{2 x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{64 cos ( 8 x )}{2})

= \frac{1}{4} \cdot x \to 0 lim (\frac{64 cos ( 8 x )}{2})

Plug in the value

x = 0

= \frac{1}{4} \cdot \frac{64 cos ( 8 \cdot 0 )}{2}

Simplify

\frac{1}{4} \cdot \frac{64 cos ( 8 \cdot 0 )}{2} : 8

= 8

\frac{\partial}{\partial x} (8 x^{9 y})

e^{- x^{3}}

\int cos (d) x d x

x \to 0 lim (\frac{1}{e ^{x} - 1} - \frac{1}{x})

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

What is the limit as x approaches 0 of (1-cos(8x))/(4x^2) ?
The limit as x approaches 0 of (1-cos(8x))/(4x^2) is 8