bemath-lim-x-0-1-3sin-x-1-3-1-sin-3x-1-3-2x-

Question Number 106614 by bemath last updated on 06/Aug/20

\underset{@ bemath @}{}

\lim_{x \to 0} \frac{\sqrt[3]{1 + 3 \sin x} - \sqrt[3]{1 + \sin 3 x}}{2 x} = ?

Answered by Dwaipayan Shikari last updated on 06/Aug/20

\lim_{x \to 0} \frac{1 + s i n x - 1 - \frac{s i n 3 x}{3}}{2 x} = \frac{x - \frac{x^{3}}{3!} - \frac{3 x}{3} + \frac{{(3 x)}^{3}}{3.3!}}{2 x} = \frac{\frac{9 x^{3}}{6} - \frac{x^{3}}{6}}{2 x} = \frac{4 x^{2}}{6} = 0

Commented by bemath last updated on 06/Aug/20

\sin x = x - \frac{x^{3}}{3!} + \dots ?

Commented by Dwaipayan Shikari last updated on 06/Aug/20

O h i e d i t e d m y e r r o r

Commented by bemath last updated on 06/Aug/20

thank you

Answered by john santu last updated on 06/Aug/20

\lim_{x \to 0} \frac{(1 + \frac{3 \sin x}{3}) - (1 + \frac{\sin 3 x}{3})}{2 x} =

\lim_{x \to 0} \frac{(x - \frac{x^{3}}{6}) - (\frac{3 x - \frac{{27 x}^{3}}{6}}{3})}{2 x} =

\lim_{x \to 0} \frac{- \frac{x^{3}}{6} + \frac{{9 x}^{3}}{6}}{2 x} = \lim_{x \to 0} \frac{{4 x}^{3}}{6 x} = 0 .

@ JS @

Answered by 1549442205PVT last updated on 06/Aug/20

Multiplying nominator and deminator by conjulate expression we get

\frac{\sqrt[3]{1 + 3 \sin x} - \sqrt[3]{1 + \sin 3 x}}{2 x}

= \frac{(1 - 3 sinx) - (1 + \sin 3 x)}{x [{(^{3} \sqrt{1 + 3 sinx})}^{2} +^{3} \sqrt{(1 + 3 sinx) (1 + \sin 3 x)} + {(^{3} \sqrt{1 + \sin 3 x})}^{2}]}

== \frac{3 sinx - \sin 3 x}{x [{(^{3} \sqrt{1 + 3 sinx})}^{2} +^{3} \sqrt{(1 + 3 sinx) (1 + \sin 3 x)} + {(^{3} \sqrt{1 + \sin 3 x})}^{2}]}

= \frac{[3 sinx - (3 sinx - {4 \sin}^{3} x]}{x [{(^{3} \sqrt{1 + 3 sinx})}^{2} +^{3} \sqrt{(1 + 3 sinx) (1 + \sin 3 x)} + {(^{3} \sqrt{1 + \sin 3 x})}^{2}]}

= \frac{sinx}{x} \times \frac{{4 \sin}^{2} x}{x [{(^{3} \sqrt{1 + 3 sinx})}^{2} +^{3} \sqrt{(1 + 3 sinx) (1 + \sin 3 x)} + {(^{3} \sqrt{1 + \sin 3 x})}^{2}]}

Hence, \lim_{x \to 0} \frac{\sqrt[3]{1 + 3 \sin x} - \sqrt[3]{1 + \sin 3 x}}{2 x} =

\lim_{x \to 0} = \frac{sinx}{x} \times \frac{{4 \sin}^{2} x}{[{(^{3} \sqrt{1 + 3 sinx})}^{2} +^{3} \sqrt{(1 + 3 sinx) (1 + \sin 3 x)} + {(^{3} \sqrt{1 + \sin 3 x})}^{2}]}

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