Question-10033

Question Number 10033 by ridwan balatif last updated on 21/Jan/17

Answered by mrW1 last updated on 21/Jan/17

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

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

= \lim_{x \to 0} \frac{x (- {2 \sin}^{2} 2 x^{2}) \cos^{2} 2 x}{- \sin^{2} 2 x}

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

= \frac{2 \times 0 \times 1 \times 1}{1} = 0

Commented by ridwan balatif last updated on 22/Jan/17

thank you sir