lim-x-0-sin-2x-p-sin-x-x-3-q-where-q-finite-

Question Number 106398 by bobhans last updated on 05/Aug/20

\lim_{x \to 0} \frac{\sin 2 x + p \sin x}{x^{3}} = q . where q finite

Answered by bemath last updated on 05/Aug/20

\lim_{x \to 0} \frac{(2 x - \frac{{8 x}^{3}}{6}) + p (x - \frac{x^{3}}{6})}{x^{3}} = q

since q finite, so 2 + p = 0, p = - 2

now we can find q

\lim_{x \to 0} \frac{- (\frac{9}{6}) x^{3}}{x^{3}} = q \Rightarrow q = - \frac{3}{2}

∴ {\begin{cases} p = - 2 \\ q = - \frac{3}{2} \end{cases}

Commented by bobhans last updated on 05/Aug/20

good

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