What is the limit as x approaches 0 of (2x^2)/(sin^2(8x)) ?

The limit as x approaches 0 of (2x^2)/(sin^2(8x)) is 1/32

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limit as x approaches 0 of (2x^2)/(sin^2(8x))

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

\frac{1}{32}

Decimal

0.03125

Solution steps

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

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

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

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2 x}{16 sin ( 8 x ) cos ( 8 x )})

= 2 \cdot x \to 0 lim (\frac{2 x}{16 sin ( 8 x ) cos ( 8 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2}{128 cos ( 16 x )})

= 2 \cdot x \to 0 lim (\frac{2}{128 cos ( 16 x )})

Plug in the value

x = 0

= 2 \cdot \frac{2}{128 cos ( 16 \cdot 0 )}

Simplify

2 \cdot \frac{2}{128 cos ( 16 \cdot 0 )} : \frac{1}{32}

= \frac{1}{32}

f (x) = x^{4} - 2 x^{3} - 5 x - 3, at x = 1

\frac{1}{16 x ^{2}}, x, \frac{x}{10}

\int (3 x - 1) (x + 5) d x

f (x) = (3 + x^{5})^{3} - (2 + x^{3})^{4}

\frac{d}{d x} (arccos (\frac{4}{x}))

What is the limit as x approaches 0 of (2x^2)/(sin^2(8x)) ?
The limit as x approaches 0 of (2x^2)/(sin^2(8x)) is 1/32