Bài 109480

Question

Tìm các giới hạn sau:
a) $\mathop {\lim }\limits_{x \to 0}\frac{2x(3x-\sin x)}{1- \cos x} $ b) $\mathop {\lim }\limits_{x \to 0}\frac{1-\cos 3x \cos 2x}{x^2} $
c) $\mathop {\lim }\limits_{x \to 5}\frac{\sin (x-5)}{3-\sqrt[]{x^2-16} } $ d) $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt[]{2+\tan x}-\sqrt[]{2+\sin x} }{x^3} $

score 0 · Answer 1

a) $\frac{2x(3x-\sin x)} {1-\cos x } =\frac{2x(3x-\sin x)}{2 \sin^2 \frac{x}{2}} =\frac{x^2\left ( 3-\frac{\sin x}{x} \right ) }{\frac{x^2}{2}\left ( \frac{\ sin \frac{x}{2} }{\frac{x}{2} } \right )^2 } =2 \frac{3- \frac{\sin x}{x} }{\left ( \frac{\sin \frac{x}{2} }{\frac{x}{2} } \right )^2 } $
Suy ra     $\mathop {\lim }\limits_{x \to 0} \frac{2x(3x-\sin x)}{1-\cos x}=2. \lim \frac{3-\frac{\sin x}{x} }{\left ( \frac{\sin \left ( \frac{x}{2} \right ) }{\frac{x}{2} } \right )^2 }=2. \frac{3-1}{1^2}=4 $
b) $\cos 3x. \cos 2x=\frac{1}{2} \left ( \cos \frac{5x}{2}+ \cos \frac{x}{2}\right ) $
$\Rightarrow \frac{1-\cos 3x \cos 2x}{x^2}=\frac{1-\frac{1}{2}\left ( \cos \frac{5x}{2} +\cos \frac{x}{2} \right ) }{x^2} $
        $\frac{1-\cos 3x. \cos 2x}{x^2}=\frac{\frac{1}{2}\left ( 1-\cos \frac{5x}{2} \right )+\left ( \frac{1}{2} -\cos \frac{x}{2}\right)}{x^2}=\frac{\sin^2 \frac{5x}{4} +\sin^2 \frac{x}{4} }{x^2} (x\neq 0) $
Vậy $\mathop {\lim }\limits_{x \to 0} \frac{1-\cos 3x \cos 2x}{x^2}=\mathop {\lim }\limits_{x \to 0}(\frac{\sin \frac{5x}{4} }{x})^2 + \mathop {\lim }\limits_{x \to 0} \left ( \frac{\sin \frac{x}{4} }{x} \right )^2=\frac{25}{16}+\frac{1}{16}=\frac{26}{16}=\frac{13}{8}        $
c)     $\frac{\sin(x-5)}{3-\sqrt[]{x^2-16} }=-\frac{\sin(x-5)(3+\sqrt[]{x^2-16} )}{(x-5)(x+5)} $
Suy ra: $\mathop {\lim }\limits_{x \to 5}\frac{\sin(x-5)}{3-\sqrt[]{x^2-16} }=-\mathop {\lim }\limits_{x \to 5}\frac{\sin (x-5)}{x-5}. \mathop {\lim }\limits_{x \to 5} \frac{(3+\sqrt[]{x^2-16} )}{x+5}=-

1.\frac{6}{10}=-\frac{3}{5}        $
d)    $\frac{\sqrt[]{2+\tan x}-\sqrt[]{2+\sin x} }{x^3}=\frac{\tan x- \sin x}{x^3(\sqrt[]{2+\tan x}+\sqrt[]{2+\sin x} )} $
                                         $=\frac{\tan x(1-\cos^2 x)}{x^3(1+\cos x)(\sqrt[]{2+\tan x}+\sqrt[]{2+\sin x}\ )} $
                                         $=\frac{\sin^3 x}{x^3}. \frac{1}{\cos x(1+\cos x)(\sqrt[]{2+\tan x}+\sqrt[]{2+\sin x} )} $
Suy ra:    $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt[]{2+\tan x}-\sqrt[]{2+\sin x} }{x^3} =\frac{1}{4\sqrt[]{2} } =\frac{\sqrt[]{2} }{8} $

Bài 109480

1 Đáp án

Lý thuyết liên quan

Bài 109480

1 Đáp án

Liên quan

Lý thuyết liên quan