What is the limit as h approaches 0 of (sin(9h))/(9h) ?

The limit as h approaches 0 of (sin(9h))/(9h) is 1

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limit as h approaches 0 of (sin(9h))/(9h)

h \to 0 lim (\frac{sin ( 9 h )}{9 h})

1

Solution steps

h \to 0 lim (\frac{sin ( 9 h )}{9 h})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{9} \cdot h \to 0 lim (\frac{sin ( 9 h )}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{cos ( 9 h ) \cdot 9}{1})

= \frac{1}{9} \cdot h \to 0 lim (\frac{cos ( 9 h ) \cdot 9}{1})

Plug in the value

h = 0

= \frac{1}{9} \cdot \frac{cos ( 9 \cdot 0 ) \cdot 9}{1}

Simplify

\frac{1}{9} \cdot \frac{cos ( 9 \cdot 0 ) \cdot 9}{1} : 1

= 1

x \to 0 lim (x + 9 - 3)

\frac{d}{d x} (\frac{x ^{2}}{x - 8})

\int \frac{x ^{2}}{x ^{2} + x + 6} d x

x + \frac{1}{x ^{3}}

x \to 2 lim (x^{2} - 4)

What is the limit as h approaches 0 of (sin(9h))/(9h) ?
The limit as h approaches 0 of (sin(9h))/(9h) is 1