What is the limit as h approaches 0 of (cos(6h)-1)/h ?

The limit as h approaches 0 of (cos(6h)-1)/h is 0

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limit as h approaches 0 of (cos(6h)-1)/h

h \to 0 lim (\frac{cos ( 6 h ) - 1}{h})

0

Solution steps

h \to 0 lim (\frac{cos ( 6 h ) - 1}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{- 6 sin ( 6 h )}{1})

= h \to 0 lim (\frac{- 6 sin ( 6 h )}{1})

Plug in the value

h = 0

= \frac{- 6 sin ( 6 \cdot 0 )}{1}

Simplify

\frac{- 6 sin ( 6 \cdot 0 )}{1} : 0

= 0

\int \frac{x ^{3} + 4 x ^{2} + x - 1}{x ^{3} + x ^{2}} d x

y^{^{''}} + y = 2 + x^{2}

\frac{d ^{2} y}{d x ^{2}} = 49 y

\frac{d}{d x} (6^{4} x^{3} + x^{5} - 180 4 x^{2})

\frac{d}{d x} (\frac{- e ^{- x} x - e ^{- x}}{x ^{2}})

What is the limit as h approaches 0 of (cos(6h)-1)/h ?
The limit as h approaches 0 of (cos(6h)-1)/h is 0