What is the limit as h approaches 0 of ((sin(pi/6+h)-1/2))/h ?

The limit as h approaches 0 of ((sin(pi/6+h)-1/2))/h is (sqrt(3))/2

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limit as h approaches 0 of ((sin(pi/6+h)-1/2))/h

h \to 0 lim (\frac{( sin ( \frac{π}{6} + h ) - \frac{1}{2} )}{h})

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

h \to 0 lim (\frac{sin ( \frac{π}{6} + h ) - \frac{1}{2}}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{cos ( h + \frac{π}{6} )}{1})

= h \to 0 lim (\frac{cos ( h + \frac{π}{6} )}{1})

Plug in the value

h = 0

= \frac{cos ( 0 + \frac{π}{6} )}{1}

Simplify

\frac{cos ( 0 + \frac{π}{6} )}{1} : \frac{3}{2}

= \frac{3}{2}

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What is the limit as h approaches 0 of ((sin(pi/6+h)-1/2))/h ?
The limit as h approaches 0 of ((sin(pi/6+h)-1/2))/h is (sqrt(3))/2