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})

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

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