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

The limit as x approaches 0 of (sin^4(2x))/(4x^4) is 4

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limit as x approaches 0 of (sin^4(2x))/(4x^4)

x \to 0 lim (\frac{sin ^{4} ( 2 x )}{4 x ^{4}})

4

Solution steps

x \to 0 lim (\frac{sin ^{4} ( 2 x )}{4 x ^{4}})

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{sin ^{4} ( 2 x )}{x ^{4}})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{8 sin ^{3} ( 2 x ) cos ( 2 x )}{4 x ^{3}})

= \frac{1}{4} \cdot x \to 0 lim (\frac{8 sin ^{3} ( 2 x ) cos ( 2 x )}{4 x ^{3}})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{8 ( 6 sin ^{2} ( 2 x ) cos ^{2} ( 2 x ) - 2 sin ^{4} ( 2 x ) )}{12 x ^{2}})

= \frac{1}{4} \cdot x \to 0 lim (\frac{8 ( 6 sin ^{2} ( 2 x ) cos ^{2} ( 2 x ) - 2 sin ^{4} ( 2 x ) )}{12 x ^{2}})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{8 ( 12 sin ( 4 x ) cos ( 4 x ) - 16 sin ^{3} ( 2 x ) cos ( 2 x ) )}{24 x})

= \frac{1}{4} \cdot x \to 0 lim (\frac{8 ( 12 sin ( 4 x ) cos ( 4 x ) - 16 sin ^{3} ( 2 x ) cos ( 2 x ) )}{24 x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{8 ( 48 cos ( 8 x ) - 16 ( 6 sin ^{2} ( 2 x ) cos ^{2} ( 2 x ) - 2 sin ^{4} ( 2 x ) ) )}{24})

= \frac{1}{4} \cdot x \to 0 lim (\frac{8 ( 48 cos ( 8 x ) - 16 ( 6 sin ^{2} ( 2 x ) cos ^{2} ( 2 x ) - 2 sin ^{4} ( 2 x ) ) )}{24})

Plug in the value

x = 0

= \frac{1}{4} \cdot \frac{8 ( 48 cos ( 8 \cdot 0 ) - 16 ( 6 sin ^{2} ( 2 \cdot 0 ) cos ^{2} ( 2 \cdot 0 ) - 2 sin ^{4} ( 2 \cdot 0 ) ) )}{24}

Simplify

\frac{1}{4} \cdot \frac{8 ( 48 cos ( 8 \cdot 0 ) - 16 ( 6 sin ^{2} ( 2 \cdot 0 ) cos ^{2} ( 2 \cdot 0 ) - 2 sin ^{4} ( 2 \cdot 0 ) ) )}{24} : 4

= 4

\frac{d v}{d t} = 0.64 - \frac{v}{2940} \cdot 8, v (0) = 0

\int (t^{3} - 3 t) (2 t) d t

\frac{d x}{d t} = 8 x t^{5}

\int sin^{2} (x) cos^{3} (x) d x

x y + y^{2} - x^{2} y^{^{'}} = 0

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