lim-x-0-x-tan-x-x-sin-x-cos-x-1-

Question Number 18587 by Joel577 last updated on 25/Jul/17

\lim_{x \to 0} \frac{x . \tan x}{x \sin x - \cos x + 1}

Commented by Joel577 last updated on 25/Jul/17

\Rightarrow \lim_{x \to 0} \frac{x . \tan x}{x \sin x + {2 \sin}^{2} \frac{1}{2} x} = \lim_{x \to 0} \frac{x . \tan x}{\sin x (x + 2 \sin \frac{1}{2} x)}

\Rightarrow \lim_{x \to 0} \frac{x}{\sin x} . \lim_{x \to 0} \frac{\tan x}{x + 2 \sin \frac{1}{2} x}

\Rightarrow 1 . \lim_{x \to 0} \frac{\frac{\tan x}{x}}{\frac{x + 2 \sin \frac{1}{2} x}{x}} = 1 . \frac{1}{1 + 1} = \frac{1}{2}

Is this correct ?

Commented by mrW1 last updated on 26/Jul/17

no, you made a mistake .

\Rightarrow \lim_{x \to 0} \frac{x . \tan x}{x \sin x + {2 \sin}^{2} \frac{1}{2} x}

\Rightarrow \lim_{x \to 0} \frac{\frac{\tan x}{x}}{\frac{\sin x}{x} + \frac{1}{2} {(\frac{\sin \frac{x}{2}}{\frac{x}{2}})}^{2}}

\Rightarrow \frac{1}{1 + \frac{1}{2} \times 1^{2}} = \frac{2}{3}

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