What is the limit as x approaches 0 of (1-cos(4x))/(1-cos(6x)) ?

The limit as x approaches 0 of (1-cos(4x))/(1-cos(6x)) is 4/9

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limit as x approaches 0 of (1-cos(4x))/(1-cos(6x))

x \to 0 lim (\frac{1 - cos ( 4 x )}{1 - cos ( 6 x )})

\frac{4}{9}

Decimal

0.44444 \dots

Solution steps

x \to 0 lim (\frac{1 - cos ( 4 x )}{1 - cos ( 6 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{4 sin ( 4 x )}{6 sin ( 6 x )})

= x \to 0 lim (\frac{4 sin ( 4 x )}{6 sin ( 6 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{16 cos ( 4 x )}{36 cos ( 6 x )})

= x \to 0 lim (\frac{16 cos ( 4 x )}{36 cos ( 6 x )})

Plug in the value

x = 0

= \frac{16 cos ( 4 \cdot 0 )}{36 cos ( 6 \cdot 0 )}

Simplify

\frac{16 cos ( 4 \cdot 0 )}{36 cos ( 6 \cdot 0 )} : \frac{4}{9}

= \frac{4}{9}

x \to 2 lim (\frac{1}{( x - 1 ) ^{4} + 2})

x \to 2 lim (- x + 1)

x \to 4 lim (\frac{x ^{2} - 3 x - 4}{2 x + 1})

x \to - 2 lim (x^{2} + a)

x \to 5 - lim (\frac{( 1 )}{x - 5})

What is the limit as x approaches 0 of (1-cos(4x))/(1-cos(6x)) ?
The limit as x approaches 0 of (1-cos(4x))/(1-cos(6x)) is 4/9