What is the limit as x approaches pi/4 of (sin(2x)-1)/(4x-pi) ?

The limit as x approaches pi/4 of (sin(2x)-1)/(4x-pi) is 0

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limit as x approaches pi/4 of (sin(2x)-1)/(4x-pi)

x \to \frac{π}{4} lim (\frac{sin ( 2 x ) - 1}{4 x - π})

0

Solution steps

x \to \frac{π}{4} lim (\frac{sin ( 2 x ) - 1}{4 x - π})

Apply L'Hopital's Rule

: x \to \frac{π}{4} lim (\frac{2 cos ( 2 x )}{4})

= x \to \frac{π}{4} lim (\frac{2 cos ( 2 x )}{4})

Plug in the value

x = \frac{π}{4}

= \frac{2 cos ( 2 \cdot \frac{π}{4} )}{4}

Simplify

\frac{2 cos ( 2 \cdot \frac{π}{4} )}{4} : 0

= 0

\frac{d}{d x} (cos (2 x + 3 y))

y^{^{''}} - 16 y^{^{'}} + 63 y = 240 e^{x}

\int \frac{x - 5}{x ^{2} + 2 x - 3} d x

f (x) = \frac{3}{5 x + 2}, at x = - 1

y = x, y = x^{3}

What is the limit as x approaches pi/4 of (sin(2x)-1)/(4x-pi) ?
The limit as x approaches pi/4 of (sin(2x)-1)/(4x-pi) is 0