lim-x-0-1-tan-1-x-sin-x-1-x-3-

Question Number 159727 by blackmamba last updated on 20/Nov/21

\lim_{x \to 0} \sqrt[x^{3}]{1 + \tan (1 - (\frac{x}{\sin x}))} ?

Answered by FongXD last updated on 20/Nov/21

L = \lim_{x \to 0^{+}} {[1 + \tan (1 - \frac{x}{sinx})]}^{\frac{1}{x^{3}}}

L = \lim_{x \to 0^{+}} {{[1 + \tan (1 - \frac{x}{sinx})]}^{\frac{1}{\tan (1 - \frac{x}{sinx})}}}^{\frac{\tan (1 - \frac{x}{sinx})}{x^{3}}}

L = {\underset{x \to 0^{+}}{\lim e}}^{\frac{\tan (1 - \frac{x}{sinx})}{x^{3}}} = e^{\lim_{x \to 0^{+}} \frac{\tan (1 - \frac{x}{sinx})}{1 - \frac{x}{sinx}} \times \frac{sinx - x}{x^{3} sinx}}

L = e^{\lim_{x \to 0^{+}} \frac{sinx - x}{x^{3}} \times \lim_{x \to 0^{+}} \frac{1}{sinx}}

where M = \lim_{x \to 0^{+}} \frac{sinx - x}{x^{3}} = \lim_{x \to 0^{+}} \frac{\sin 3 x - 3 x}{{27 x}^{3}} (change x to 3 x)

\Leftrightarrow 27 M = \lim_{x \to 0^{+}} \frac{3 sinx - {4 \sin}^{3} x - 3 x}{x^{3}} = 3 \lim_{x \to 0^{+}} \frac{sinx - x}{x^{3}} - 4 \lim_{x \to 0^{+}} {(\frac{sinx}{x})}^{3}

\Leftrightarrow 27 M = 3 M - 4, \Rightarrow M = \lim_{x \to 0^{+}} \frac{sinx - x}{x^{3}} = - \frac{1}{6}

then L = e^{- \frac{1}{6} \lim_{x \to 0^{+}} \frac{1}{sinx}} = 0

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