Bài 112909

Question

Tính giới hạn :
a) $\mathop {\lim }\limits_{x \to 0} \frac{1 - \sqrt{\cos x}}{1 - \cos \sqrt{x} }$ b) $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{1 - \tan x}-\sqrt{1+ \sin x}}{x^3}$

score 0 · Answer 1

a) Ta có :
      $ \mathop {\lim }\limits_{x \to 0} \frac{1 -\sqrt{\cos x} }{1 - \cos \sqrt{x} } = \mathop {\lim }\limits_{x \to 0} \frac{1 - \cos x}{(1-\cos \sqrt{x} )(1+\sqrt{\cos x} )}$
                                 $ = \mathop {\lim }\limits_{x \to 0} \frac{2 \sin ^2 \frac{x}{2} }{2 \sin ^2 \frac{\sqrt{x} }{2}(1+\sqrt{\cos x} ) }$
                                 $ = \mathop {\lim }\limits_{x \to 0} \left [ \frac{\sin ^2 \frac{x}{2} }{\left ( \frac{x}{2} \right )^2 }. \left ( \frac{x}{2} \right )^2.\frac{\left ( \frac{\sqrt{x} }{2} \right )^2 }{\sin ^2 . \frac{\sqrt{x} }{2} }.\frac{1}{\left ( \frac{\sqrt{x} }{2} \right )^2 }.\frac{1}{1 +\sqrt{\cos x} } \right ]$
                                 $=\mathop {\lim }\limits_{x \to 0} \frac{x}{1+\sqrt{\cos x}} = 0$.
b) Ta có :
      $ \mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x} }{x^3} = \mathop {\lim }\limits_{x \to 0} \frac{\tan x - \sin x}{x^3(\sqrt{1 + \tan x}+\sqrt{1+ \sin x})}$
                                          $ = \mathop {\lim }\limits_{x \to 0} \frac{\tan x(1-\cos x)}{x^3(\sqrt{1 + \tan x}+ \sqrt{1 + \sin x})   } $
                                          $ = \mathop {\lim }\limits_{x \to 0} \frac{2 \tan x . \sin^2 \frac{x}{2} }{x^3 (\sqrt{1 + \tan x}+ \sqrt{1 + \sin x}) } $
                                          $ = 2 \mathop {\lim }\limits_{x \to 0} \left [ \frac{\tan x}{x}\frac{\sin ^2 \frac{x}{2} }{\left ( \frac{x}{2} \right )^2.4 }.\frac{1}{\sqrt{1+\tan x}+\sqrt{1+ \sin x} }    \right ] =\frac{1}{4}$

Bài 112909

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Bài 112909

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