Bài 112900

Question

Tính giới hạn :
a) $\mathop {\lim }\limits_{x \to a}\frac{\sqrt{x} -\sqrt{a}+\sqrt{x-a} }{\sqrt{x^2-a^2} }$, với $a>0$
b) $\mathop {\lim }\limits_{x \to 1}\frac{x-1}{\sqrt{x^2+3}+x^3-3x }$.

score 0 · Answer 1

a) Ta có :
      $\mathop {\lim }\limits_{x \to a^+}\frac{\sqrt{x} -\sqrt{a}+\sqrt{x-a} }{\sqrt{x^2-a^2} } = \mathop {\lim }\limits_{x \to a^+}\left ( \frac{\sqrt{x} -\sqrt{a} }{\sqrt{x^2-a^2} } + \frac{\sqrt{x-a} }{\sqrt{x^2-a^2} } \right )$
                                 $= \mathop {\lim }\limits_{x \to a^+} \left ( \frac{x-a}{(\sqrt{x} +\sqrt{a})\sqrt{x^2-a^2} } +\frac{1}{\sqrt{x+a} } \right ) $
                                 $= \mathop {\lim }\limits_{x \to a^+} \left ( \frac{\sqrt{x-a} }{(\sqrt{x} +\sqrt{a})\sqrt{x+a} }+\frac{1}{\sqrt{2a} } \right ) = \frac{1}{\sqrt{2a} }$.
b) Ta có :
      $\mathop {\lim }\limits_{x \to 1}\frac{x-1}{\sqrt{x^2+3}+x^3-3x } = \mathop {\lim }\limits_{x \to 1}\frac{1}{\frac{\sqrt{x^2+3}-2 }{x-1}+\frac{x^3-3x+2}{x-1} }$
                                               $= \mathop {\lim }\limits_{x \to 1}\frac{1}{\frac{x^2-1}{(\sqrt{x^2+3}+2)(x-1) }+x-2 }$
                                               $= \mathop {\lim }\limits_{x \to 1}\frac{1}{\frac{x+1}{\sqrt{x^2+3}+2}+x-2 } = -2$.

Bài 112900

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Bài 112900

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