Bài 107850

Question

Xác định dạng vô định và tính các giới hạn sau:
a) $\mathop {\lim }\limits_{x \to 2}\frac{x^2-x-2}{3x^2-5x-2} $             b) $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt[]{x^2+1}-1 }{x^2}$
c) $\mathop {\lim }\limits_{x \to +\infty} \frac{-x^3-x^2+2}{3x^3+4x+6} $                                       d) $\mathop {\lim }\limits_{x \to -\infty}\frac{3x+1}{\sqrt[]{x(x-1)}+\sqrt[]{9x^2+2}}$
e) $\mathop {\lim }\limits_{x \to 0^-}\frac{2}{x}(\frac{3}{x+1}-3) $                                    f) $\mathop {\lim }\limits_{x \to -\infty} (\sqrt[]{9x^2-x}+3x ) $.

score 0 · Answer 1

a) Dạng $\frac{0}{0} $:   $\mathop {\lim }\limits_{x \to 2}\frac{x^2-x-2}{3x^2-5x-2}=\mathop {\lim }\limits_{x \to 2}

\frac{(x-2)(x+1)}{3(x-2)(x+\frac{1}{3} )}=\mathop {\lim }\limits_{x \to 2}\frac{x+1}{3x+1} =\frac{3}{7} $

b) Dạng $\frac{0}{0}$:   $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt[]{x^2+1}-1}{x^2}= \mathop {\lim }\limits_{x \to

0} \frac{(x^2+1)-1}{x^2(\sqrt[]{x^2+1}+1)}= \mathop {\lim }\limits_{x \to 0}\frac{1}{\sqrt[]{x^2+1}+1 }=\frac{1}{2}$

c) Dạng $\frac{\infty }{\infty}: \mathop {\lim }\limits_{x \to +\infty}\frac{-x^3-x^2+2}{3x^3+4x+6}=\mathop {\lim

}\limits_{x \to \infty} \dfrac{-1-\frac{1}{x^2}+\frac{2}{x^3}}{3+\frac{4}{x^2}+\frac{6}{x^3}}=-\frac{1}{3}      $

d) Dạng $\frac{\infty}{\infty} $: $\mathop {\lim }\limits_{x \to -\infty}\frac{3x+1}{\sqrt[]{x(x-1)}+\sqrt[]{9x^2+2}}

=\mathop {\lim }\limits_{x \to -\infty}\frac{3x+1}{|x|\left ( \sqrt[]{1-\frac{1}{x} } \right ) +3|x| \left ( \sqrt[]{1+\frac{2}{9x^2}} \right ) } $
                         $=\mathop {\lim }\limits_{x \to -\infty}\frac{3+\frac{1}{x} }{-\sqrt[]{1-\frac{1}{x} }-\sqrt[]{9+\frac{2}{x^2}}}=-\frac{3}

{4}$

e) Dạng $0.\infty$:   $\mathop {\lim }\limits_{x \to 0^-}\frac{2}{x}(\frac{3}{x+1}-3 )=\mathop {\lim }\limits_{x \to 0}

\frac{2}{x}.(-\frac{3x}{x+1} )= \mathop {\lim }\limits_{x \to 0^-}\frac{-6}{x+1}=-6      $

f) Dạng $\infty-\infty $: $ \mathop {\lim }\limits_{x \to -\infty} (\sqrt[]{9x^2-x}+3x )=\mathop {\lim }\limits_{x \to -

\infty} \frac{9x^2-x-9x^2}{\sqrt[]{9x^2-x}-3x} $
                          $=\mathop {\lim }\limits_{x \to -\infty} \frac{-x}{|x|\sqrt[]{9-\frac{1}{x}}

-3x}=\mathop {\lim }\limits_{x \to -\infty} \frac{-x}{-x\sqrt[]{9-\frac{1}{x} }-3x}$
                          $ = \mathop {\lim }\limits_{x \to -

\infty} \frac{1}{\sqrt[]{9-\frac{1}{x} }+3 }=\frac{1}{6}.           $

Bài 107850

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Bài 107850

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