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

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

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limit as h approaches 0 of (1-cos(2h))/h

h \to 0 lim (\frac{1 - cos ( 2 h )}{h})

0

Solution steps

h \to 0 lim (\frac{1 - cos ( 2 h )}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{2 sin ( 2 h )}{1})

= h \to 0 lim (\frac{2 sin ( 2 h )}{1})

Plug in the value

h = 0

= \frac{2 sin ( 2 \cdot 0 )}{1}

Simplify

\frac{2 sin ( 2 \cdot 0 )}{1} : 0

= 0

x e^{- x + 1}, 1

x^{x} (ln (x) + 1)

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

4 cos^{2} (x)

1 9^{x}

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