"Finding the limit of1−cos⁡(2x)1−cos⁡(3x) for x→0" was answered by students like you

fecundavai3c

Answered question

2022-03-01

Finding the limit of

\frac{1 - \cos (2 x)}{1 - \cos (3 x)}

for

x \to 0

mtakadamu9i5

Beginner2022-03-02Added 8 answers

Putting

x = 2 y

lim_{x \to 0} \frac{1 - \cos (2 x)}{1 - \cos (3 x)}

= lim_{y \to 0} \frac{1 - \cos 4 y}{1 - \cos 6 y}

= lim_{y \to 0} \frac{2 \sin^{2} 2 y}{2 \sin^{2} 3 y}

\cos 2 θ = 1 - 2 \sin^{2} θ

= lim_{y \to 0} \frac{4 \cdot {(\frac{\sin 2 y}{2 y})}^{2}}{9 \cdot {(\frac{\sin 3 y}{3 y})}^{2}}

= \frac{4}{9}

lim_{h \to 0} \frac{\sin h}{h} = 1

Alternatively,

\frac{1 - \cos (2 x)}{1 - \cos 3 x} = {(\frac{\sin 2 x}{\sin 3 x})}^{2} \cdot (\frac{1 + \cos 3 x}{1 + \cos 2 x})

Now

lim_{x \to 0} \cos a x = 1

for finite a

\Rightarrow lim_{x \to 0} (\frac{1 + \cos 3 x}{1 + \cos 2 x}) = \frac{1 + 1}{1 + 1} = 1

and

lim_{x \to 0} (\frac{\sin 2 x}{\sin 3 x}) = \frac{2}{3} \cdot \frac{lim_{x \to 0 \frac{\sin 2 x}{2 x}}}{lim_{x \to 0} \frac{\sin 3 x}{3 x}} = \frac{2}{3}

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