Đáp án:
b) \( - \dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\lim n.\dfrac{{2{n^3} - 3{n^2} + 4 - \dfrac{8}{n}}}{{ - 3{n^3} + {n^2} - 4}}\\
= \lim n.\dfrac{{2 - \dfrac{3}{n} + \dfrac{4}{{{n^3}}} - \dfrac{8}{{{n^4}}}}}{{ - 3 + \dfrac{1}{n} - \dfrac{4}{{{n^3}}}}} = - \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } n = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2 - \dfrac{3}{n} + \dfrac{4}{{{n^3}}} - \dfrac{8}{{{n^4}}}}}{{ - 3 + \dfrac{1}{n} - \dfrac{4}{{{n^3}}}}} = - \dfrac{2}{3}\\
b)\mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} - 3x + 1 - {x^2}}}{{\sqrt {{x^2} - 3x + 1} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - 3x + 1}}{{\sqrt {{x^2} - 3x + 1} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - 3 + \dfrac{1}{x}}}{{\sqrt {1 - \dfrac{3}{x} + \dfrac{1}{{{x^2}}}} + 1}} = - \dfrac{3}{2}\\
c)\mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left| {x - 3} \right|}}{{5\left( {x - 3} \right)}} = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{x - 3}}{{5\left( {x - 3} \right)}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{1}{5} = \dfrac{1}{5}
\end{array}\)
