What is the limit as x approaches 0 of (1-cos(7x))/(4x^2) ?

The limit as x approaches 0 of (1-cos(7x))/(4x^2) is 49/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as x approaches 0 of (1-cos(7x))/(4x^2)

x \to 0 lim (\frac{1 - cos ( 7 x )}{4 x ^{2}})

\frac{49}{8}

Decimal

6.125

Solution steps

x \to 0 lim (\frac{1 - cos ( 7 x )}{4 x ^{2}})

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

= \frac{1}{4} \cdot x \to 0 lim (\frac{1 - cos ( 7 x )}{x ^{2}})

Apply L'Hopital's Rule

= \frac{1}{4} \cdot x \to 0 lim (\frac{7 sin ( 7 x )}{2 x})

Apply L'Hopital's Rule

= \frac{1}{4} \cdot x \to 0 lim (\frac{49 cos ( 7 x )}{2})

Plug in the value

x = 0

= \frac{1}{4} \cdot \frac{49 cos ( 7 \cdot 0 )}{2}

Simplify

\frac{1}{4} \cdot \frac{49 cos ( 7 \cdot 0 )}{2} : \frac{49}{8}

= \frac{49}{8}

What is the limit as x approaches 0 of (1-cos(7x))/(4x^2) ?
The limit as x approaches 0 of (1-cos(7x))/(4x^2) is 49/8