What is the limit as h approaches 0 of (cos((3pi)/2+h)-cos((3pi)/2))/(h) ?

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

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limit as h approaches 0 of (cos((3pi)/2+h)-cos((3pi)/2))/(h)

h \to 0 lim (\frac{cos ( \frac{3 π}{2} + h ) - cos ( \frac{3 π}{2} )}{h})

1

Solution steps

h \to 0 lim (\frac{cos ( \frac{3 π}{2} + h ) - cos ( \frac{3 π}{2} )}{h})

cos (\frac{3 π}{2} + h) - cos (\frac{3 π}{2}) = cos (h + \frac{3 π}{2})

= h \to 0 lim (\frac{cos ( h + \frac{3 π}{2} )}{h})

Apply L'Hopital's Rule

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

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

Plug in the value

h = 0

= \frac{cos ( 0 )}{1}

Simplify

\frac{cos ( 0 )}{1} : 1

= 1

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