What is the limit as h approaches 0 of ((4sin(pi/2+h)-4sin(pi/2)))/h ?

The limit as h approaches 0 of ((4sin(pi/2+h)-4sin(pi/2)))/h is 0

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limit as h approaches 0 of ((4sin(pi/2+h)-4sin(pi/2)))/h

h \to 0 lim (\frac{( 4 sin ( \frac{π}{2} + h ) - 4 sin ( \frac{π}{2} ) )}{h})

0

Solution steps

h \to 0 lim (\frac{4 sin ( \frac{π}{2} + h ) - 4 sin ( \frac{π}{2} )}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{- 4 sin ( h )}{1})

= h \to 0 lim (\frac{- 4 sin ( h )}{1})

Plug in the value

h = 0

= \frac{- 4 sin ( 0 )}{1}

Simplify

\frac{- 4 sin ( 0 )}{1} : 0

= 0

x \to 0 lim (\frac{x ^{2} + 4 x + 5 + 5}{x})

x \to 0 lim (\frac{[ x ^{2} ] - x ^{2}}{x ^{2}})

x \to \infty lim (\frac{4 x}{3 x - 2})

x \to 0 lim (\frac{( 2 ^{x} - 5 ^{x} )}{x})

x \to 1 + lim (\frac{1}{x ^{2} - 3 x + 2})

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