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

The limit as x approaches 0 of (sec(2x)-1)/(7x) is 0

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limit as x approaches 0 of (sec(2x)-1)/(7x)

x \to 0 lim (\frac{sec ( 2 x ) - 1}{7 x})

0

Solution steps

x \to 0 lim (\frac{sec ( 2 x ) - 1}{7 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{sec ( 2 x ) - 1}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2 sec ( 2 x ) tan ( 2 x )}{1})

= \frac{1}{7} \cdot x \to 0 lim (\frac{2 sec ( 2 x ) tan ( 2 x )}{1})

Plug in the value

x = 0

= \frac{1}{7} \cdot \frac{2 sec ( 2 \cdot 0 ) tan ( 2 \cdot 0 )}{1}

Simplify

\frac{1}{7} \cdot \frac{2 sec ( 2 \cdot 0 ) tan ( 2 \cdot 0 )}{1} : 0

= 0

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What is the limit as x approaches 0 of (sec(2x)-1)/(7x) ?
The limit as x approaches 0 of (sec(2x)-1)/(7x) is 0