Đáp án:
\(\begin{array}{l}
a.\frac{{\sqrt 2 }}{2}\\
b.0\\
c.\frac{{12}}{7}\\
d.2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\lim \frac{{\sqrt {2 + \frac{3}{{{n^3}}} - \frac{2}{{{n^4}}}} }}{{2 - \frac{1}{n} + \frac{3}{{{n^2}}}}}\\
= \lim \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}\\
b.\lim \frac{{{n^3} + 1 - {n^3}}}{{{n^2} - n\sqrt[3]{{1 - {n^3}}} + \sqrt[3]{{{{\left( {1 - {n^3}} \right)}^2}}}}}\\
= \lim \frac{1}{{{n^2} - n\sqrt[3]{{1 - {n^3}}} + \sqrt[3]{{{{\left( {1 - {n^3}} \right)}^2}}}}}\\
= \lim \frac{{\frac{1}{{{n^2}}}}}{{1 - 1.\sqrt[3]{{\frac{1}{{{n^3}}} - 1}} + \sqrt[3]{{{{\left( {\frac{1}{{{n^3}}} - 1} \right)}^2}}}}}\\
= \lim \frac{0}{{1 + 1 + 1}} = 0\\
c.\mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {x + 9} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - 2x + 4}}{{x + 9}}\\
= \frac{{4 + 4 + 4}}{{ - 2 + 9}} = \frac{{12}}{7}\\
d.\mathop {\lim }\limits_{x \to - \infty } \frac{{2\left| x \right| + 3}}{{\sqrt {{x^2} + x + 5} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2x + 3}}{{\sqrt {{x^2} + x + 5} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2 + \frac{3}{x}}}{{ - \sqrt {1 + \frac{1}{x} + \frac{5}{{{x^2}}}} }} = 2
\end{array}\)
