Question-196959

Question Number 196959 by cortano12 last updated on 05/Sep/23

\underset{―}{}

Answered by horsebrand11 last updated on 05/Sep/23

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

= - \frac{8}{6} + 0 = 1 - b

\Rightarrow {\begin{cases} a = - 2 \\ b = \frac{7}{3} \end{cases}

Answered by MM42 last updated on 05/Sep/23

{l i m}_{x \to 0} \frac{s i n 2 x + a x + {b x}^{3}}{x^{3}}

= {l i m}_{x \to 0} \frac{2 x - \frac{4}{3} x^{3} + o (x) + a x + {b x}^{3}}{x^{3}}

= {l i m}_{x \to 0} \frac{(2 + a) x + (b - \frac{4}{3}) x^{3}}{x^{3}} = 1

\Rightarrow {\begin{cases} 2 + a = 0 \Rightarrow a = - 2 \\ b - \frac{4}{3} = 1 \Rightarrow b = \frac{7}{3} \end{cases}

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