Chủ đề này có 3 trả lời

[Winter Is Coming]

tính các giới hạn

\[\begin{gathered}
  \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {5 - {x^2}}  + \sqrt[3]{{{x^2} + 7}}}}{{{x^2} - 1}} \hfill \\
  \mathop {\lim }\limits_{x \to 0} \frac{{2\sqrt {1 + x}  - \sqrt[3]{{8 - x}}}}{x} \hfill \\
  \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {8{x^3} + {x^2} + 6x + 9}  - \sqrt[3]{{9{x^2} + 27x + 27}}}}{{{x^3}}} \hfill \\
  \mathop {\lim }\limits_{x \to 0} \frac{{{x^3}}}{{\sqrt {\left( {1 + 2x} \right)\left( {1 + {x^2}} \right)}  - \sqrt {\left( {1 + 3x} \right)\left( {1 + 3{x^2}} \right)} }} \hfill \\
\end{gathered} \]

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[Winter Is Coming]

câu 1 sai đề nha. trên tử là căn - căn nha mn. e ghi để sai

[DOTOANNANG]

$\it{3}$

Do $\it{x}\rightarrow \it{0}\Rightarrow \it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9},\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}> \it{0}$$,$ nên$:$

$\lim\limits_{\,\it{x}\rightarrow \it{0}\,}\,\frac{\sqrt{\, \it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}}- \sqrt[\it{3}\,]{\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}}}{\it{x}^{\,\it{3}}}=$

$= \lim\limits_{\,\it{x}\rightarrow \it{0}\,}\,\frac{\sqrt[\it{6}\,]{\,\it{(}\,\,\it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}\,\,\it{)}^{\,\it{3}}}- \sqrt[\it{6}\,]{\,\it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}}}{\it{(}\,\,\it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}\,\,\it{)}^{\,\it{3}}- \it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}}\,\it{(}\,\,\it{512}\,\it{x}^{\,\it{6}}+ \it{192}\,\it{x}^{\,\it{5}}+ \it{1176}\,\it{x}^{\,\it{4}}+ \it{2017}\,\it{x}^{\,\it{3}}+ \it{1314}\,\it{x}^{\,\it{2}}+ \it{2646}\,\it{x}+ \it{1998}\,\,\it{)}=$

$= \it{1998}\,\lim\limits_{\,\it{(}\,\,\it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}\,\,\it{)}^{\,\it{3}}\rightarrow \it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}= \it{constant}\,}\,\frac{\sqrt[\it{6}\,]{\,\it{(}\,\,\it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}\,\,\it{)}^{\,\it{3}}}- \sqrt[\it{6}\,]{\,\it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}}}{\it{(}\,\,\it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}\,\,\it{)}^{\,\it{3}}- \it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}}=$ $($với $\it{constant}= \it{9}^{\,\it{3}}= \it{27}^{\,\it{2}}$$)$$,$ nên $:$

$= \it{1998}\,\lim\limits_{\,\it{(}\,\,\it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}\,\,\it{)}^{\,\it{3}}\rightarrow \it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}= \it{constant}\,}\,\frac{\sqrt[\it{6}\,]{\,\it{(}\,\,\it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}\,\,\it{)}^{\,\it{3}}}- \sqrt[\it{6}\,]{\,\it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}}}{\it{(}\,\,\it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}\,\,\it{)}^{\,\it{3}}- \it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}}=$ $\it{1998}\,{\it{f}}'\it{(}\,\,\it{(}\,\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}\,\,\it{)}^{\,\it{2}}\,\,\it{)}$$,$ với $\it{f}\it{(}\,\,\it{X}\,\,\it{)}= \sqrt[\it{6}\,]{\,\it{X}}$$,$ nên$:$

$\lim\limits_{\,\it{x}\rightarrow \it{0}\,}\,\frac{\sqrt{\, \it{8}\,\it{x}^{\,\it{3}}+ \it{x}^{\,\it{2}}+ \it{6}\,\it{x}+ \it{9}}- \sqrt[\it{3}\,]{\,\it{9}\,\it{x}^{\,\it{2}}+ \it{27}\,\it{x}+ \it{27}}}{\it{x}^{\,\it{3}}}= \it{1998}\,\frac{\it{1}}{\it{6}\,\it{(}\,\,\it{27}^{\,\it{2}}\,\,\it{)}^{\,\frac{\it{5}}{\it{6}}}}= \frac{\it{37}}{\it{27}}$$.$

[DOTOANNANG]

$\it{4}$

$\lim\limits_{\,\it{x}\rightarrow \it{0}\,}\,\frac{\it{x}^{\,\it{3}}}{\sqrt{\,\it{(}\,\,\it{1}+ \it{2}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{x}^{\,\it{2}}\,\,\it{)}}- \sqrt{\,\it{(}\,\,\it{1}+ \it{3}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{3}\,\it{x}^{\,\it{2}}\,\,\it{)}}}=$

$= \lim\limits_{\,\it{x}\rightarrow \it{0}\,}\,\frac{\it{(}\,\,\it{1}+ \it{2}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{x}^{\,\it{2}}\,\,\it{)}- \it{(}\,\,\it{1}+ \it{3}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{3}\,\it{x}^{\,\it{2}}\,\,\it{)}}{\sqrt{\,\it{(}\,\,\it{1}+ \it{2}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{x}^{\,\it{2}}\,\,\it{)}}- \sqrt{\,\it{(}\,\,\it{1}+ \it{3}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{3}\,\it{x}^{\,\it{2}}\,\,\it{)}}}\,.\,\frac{\it{x}^{\,\it{2}}}{-\it{(}\,\,\it{7}\,\it{x}^{\,\it{2}}+ \it{2}\,\it{x}+ \it{1}\,\,\it{)}}= $$,$ do$:$

$\lim\limits_{\,\it{x}\rightarrow \it{0}\,}\,\frac{\it{(}\,\,\it{1}+ \it{2}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{x}^{\,\it{2}}\,\,\it{)}- \it{(}\,\,\it{1}+ \it{3}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{3}\,\it{x}^{\,\it{2}}\,\,\it{)}}{\sqrt{\,\it{(}\,\,\it{1}+ \it{2}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{x}^{\,\it{2}}\,\,\it{)}}- \sqrt{\,\it{(}\,\,\it{1}+ \it{3}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{3}\,\it{x}^{\,\it{2}}\,\,\it{)}}}= \it{constant}$$($theo bài trên$)$$,$ nên$:$ 

$$\lim\limits_{\,\it{x}\rightarrow \it{0}\,}\,\frac{\it{x}^{\,\it{3}}}{\sqrt{\,\it{(}\,\,\it{1}+ \it{2}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{x}^{\,\it{2}}\,\,\it{)}}- \sqrt{\,\it{(}\,\,\it{1}+ \it{3}\,\it{x}\,\,\it{)}\it{(}\,\,\it{1}+ \it{3}\,\it{x}^{\,\it{2}}\,\,\it{)}}}= \it{0}$$