What is the limit as h approaches 0 of (sin(h))/(7h) ?

The limit as h approaches 0 of (sin(h))/(7h) is 1/7

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limit as h approaches 0 of (sin(h))/(7h)

h \to 0 lim (\frac{sin ( h )}{7 h})

\frac{1}{7}

Decimal Notation

0.14285 \dots

Solution steps

h \to 0 lim (\frac{sin ( h )}{7 h})

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

= \frac{1}{7} \cdot h \to 0 lim (\frac{sin ( h )}{h})

Apply L'Hopital's Rule

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

= \frac{1}{7} \cdot h \to 0 lim (\frac{cos ( h )}{1})

Plug in the value

h = 0

= \frac{1}{7} \cdot \frac{cos ( 0 )}{1}

Simplify

\frac{1}{7} \cdot \frac{cos ( 0 )}{1} : \frac{1}{7}

= \frac{1}{7}

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

\int_{3}^{\infty} \frac{1}{x ^{2} - 1} d x

\int 4 \frac{sin ( 2 x )}{sin ( x )} d x

\frac{s - 6}{s + 2}

\int_{0}^{\frac{π}{3}} 81 sin^{2} (x) d x

What is the limit as h approaches 0 of (sin(h))/(7h) ?
The limit as h approaches 0 of (sin(h))/(7h) is 1/7