Bài 107922

Question

Tính các giới hạn:
a) $\mathop {\lim }\limits_{x \to -\infty}(\sqrt[]{x^2-x+3}) b) \mathop {\lim }\limits_{x \to \infty}x(\sqrt[]

{4x^2+3}-2x) $
c) $\mathop {\lim }\limits_{x \to +\infty}\frac{x^2+2}{cos^2x} d*) \mathop {\lim }\limits_{x \to +\infty}\frac

{3x^2+5sinx}{x^4-1}$

score 0 · Answer 1

a) $\mathop {\lim }\limits_{x \to -\infty}(\sqrt[]{x^2-x+3}+x )=\mathop {\lim }\limits_{x \to -\infty}\dfrac{3-x}{\frac

{\sqrt[]{x^2-x+3} }{x} -1}   $
      Vì với $x<0$ thì $x=-\sqrt[]{x^2}$
      Do đó $\mathop {\lim }\limits_{x \to -\infty} \frac{\sqrt[]{x^2-x+3} }{x}=\mathop {\lim }\limits_{x \to -\infty }\frac{\sqrt[]{x^2-x+3} }{-\sqrt[]{x^2} }=-1$
   nên $\mathop {\lim }\limits_{x \to -\infty}\dfrac{\frac{3}{x}-1 }{\frac{\sqrt[]{x^2-x+3} }{x}-1 } =\frac{-1}{-2}=\frac{1}{2}

$

b) $\mathop {\lim }\limits_{x \to +\infty }x\left ( \sqrt[]{4x^2+3}-2x \right )=\mathop {\lim }\limits_{x \to +\infty}

\frac{3x}{\sqrt[]{4x^2+3}+2x }=\mathop {\lim }\limits_{x \to +\infty}\frac{3}{\sqrt[]{4+\frac{3}{x^2} }+2 }=\frac{3}

{4}$

c) $\mathop {\lim }\limits_{x \to +\infty}\frac{x^2+2}{\cos^2x}=\mathop {\lim }\limits_{x \to +\infty}\frac{1}{\frac

{\cos^2x}{x^2+2} }    $
      Vì $0< \frac{\cos^2x}{x^2+2} \leq \frac{1}{x^2+2} $ với mọi $x$ và $\mathop {\lim }\limits_{x \to +\infty }\frac{1}{x^2+2}=0 $
      nên $\mathop {\lim }\limits_{x \to +\infty } \frac{\cos^2x}{x^2+2}=0 $
Do đó $\mathop {\lim }\limits_{x \to +\infty} \frac{x^2+2}{\cos^2x}=\mathop {\lim }\limits_{x \to +\infty }\frac{1}{\frac{\cos^2x}{x+2} }=+\infty      $

d) Vì $\frac{3x^2-5}{x^4-1} \leq \frac{3x^2+5\sin x }{x^4-1} \leq \frac{3x^2+5}{x^4-1}   $ với $x>1$

        $\mathop {\lim }\limits_{x \to +\infty }\frac{3x^2-5}{x^4-1}=\mathop {\lim }\limits_{x \to +\infty } \dfrac{\frac{3}{x^2} -\frac{5}{x^4} }{1-\frac{1}{x^4} }=0   $ và $\mathop {\lim }\limits_{x \to +\infty }\frac{3x^2+5}{x^4-1}\mathop {\lim }\limits_{x \to +\infty } \dfrac{\frac{3}{x^2} +\frac{5}{x^4} }{1-\frac{1}{x^4} }=0 $
          Suy ra $\mathop {\lim }\limits_{x \to +\infty } \frac{3x^2+5\sin x}{x^4-1}=0 $

Bài 107922

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Bài 107922

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