Question-9929

Question Number 9929 by ridwan balatif last updated on 16/Jan/17

Answered by mrW1 last updated on 18/Jan/17

\lim_{x \to 0} \frac{\sin 4 x \cdot \tan^{2} 3 x + 6 x^{2}}{2 x^{2} + \sin 3 x \cdot \cos 2 x}

= \lim_{x \to 0} \frac{4 (\frac{\sin 4 x}{4 x}) \tan^{2} 3 x + 6 x}{2 x + 3 (\frac{\sin 3 x}{3 x}) \cos 2 x}

= \frac{4 \times 1 \times 0 + 0}{0 + 3 \times 1 \times 1} = \frac{0}{3}

= 0

\Rightarrow A n s w e r A

Commented by ridwan balatif last updated on 18/Jan/17

thank you sir