What is the limit as x approaches 0 of ((1-cos(x)))/(7x*sin(x)) ?

The limit as x approaches 0 of ((1-cos(x)))/(7x*sin(x)) is 1/14

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limit as x approaches 0 of ((1-cos(x)))/(7x*sin(x))

x \to 0 lim (\frac{( 1 - cos ( x ) )}{7 x \cdot sin ( x )})

\frac{1}{14}

Decimal Notation

0.07142 \dots

Solution steps

x \to 0 lim (\frac{1 - cos ( x )}{7 x sin ( x )})

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

= \frac{1}{7} \cdot x \to 0 lim (\frac{1 - cos ( x )}{x sin ( x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{sin ( x )}{sin ( x ) + x cos ( x )})

= \frac{1}{7} \cdot x \to 0 lim (\frac{sin ( x )}{sin ( x ) + x cos ( x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{cos ( x )}{2 cos ( x ) - x sin ( x )})

= \frac{1}{7} \cdot x \to 0 lim (\frac{cos ( x )}{2 cos ( x ) - x sin ( x )})

Plug in the value

x = 0

= \frac{1}{7} \cdot \frac{cos ( 0 )}{2 cos ( 0 ) - 0 \cdot sin ( 0 )}

Simplify

\frac{1}{7} \cdot \frac{cos ( 0 )}{2 cos ( 0 ) - 0 \cdot sin ( 0 )} : \frac{1}{14}

= \frac{1}{14}

x \to 0 lim (\frac{3 x + 1}{2 x ^{2}})

x \to 0 + lim (4 (csc (x) - \frac{1}{x}))

x \to 0 lim (\frac{arccos ( x )}{x})

x \to 3 + lim (\frac{3 x - 10}{3 - x})

x \to 2 lim (2 + ∣ 2 x - 4 ∣)

What is the limit as x approaches 0 of ((1-cos(x)))/(7x*sin(x)) ?
The limit as x approaches 0 of ((1-cos(x)))/(7x*sin(x)) is 1/14