Find-lim-x-0-1-cos4x-x-2-

Question Number 203288 by hardmath last updated on 14/Jan/24

Find : \lim_{x \to 0} \frac{1 - \cos 4 x}{x^{2}} = ?

Answered by MM42 last updated on 14/Jan/24

= {l i m}_{x \to 0} \frac{2 {s i n}^{2} 2 x}{x^{2}}

= {l i m}_{x \to 0} 8 {(\frac{s i n 2 x}{2 x})}^{2} = 8 ✓

Commented by hardmath last updated on 14/Jan/24

thankyou dearprofessor

Answered by Calculusboy last updated on 15/Jan/24

S o l u t i o n : N B t h a t l i \underset{x \to 0}{m} \frac{1 - c o s a x}{x^{2}} = \frac{a^{2}}{2}

t h e n; l i \underset{x \to 0}{m} \frac{1 - c o s 4 x}{x^{2}} = \frac{4^{2}}{2} = \frac{16}{2} = 8

M f

Answered by dimentri last updated on 15/Jan/24

= \lim_{x \to 0} \frac{1 - \cos^{2} 4 x}{x^{2} (1 + \cos 4 x)}

= \lim_{x \to 0} {(\frac{\sin 4 x}{x})}^{2} . \lim_{x \to 0} \frac{1}{1 + \cos 4 x}

= 16 \times \frac{1}{2} = 8

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