Tính giới hạn hàm số căn 9x⁴+x+1 +3x khi x -> -vô cung 19/11/2021 Bởi Allison Tính giới hạn hàm số căn 9x⁴+x+1 +3x khi x -> -vô cung
Giải thích các bước giải: Ta có: \(\begin{array}{l}\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {9{x^2} + x + 1} + 3x} \right)\\ = \mathop {\lim }\limits_{x \to – \infty } \dfrac{{\left( {\sqrt {9{x^2} + x + 1} + 3x} \right)\left( {\sqrt {9{x^2} + x + 1} – 3x} \right)}}{{\sqrt {9{x^2} + x + 1} – 3x}}\\ = \mathop {\lim }\limits_{x \to – \infty } \dfrac{{\left( {9{x^2} + x + 1} \right) – {{\left( {3x} \right)}^2}}}{{\sqrt {{x^2}\left( {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} \right)} – 3x}}\\ = \mathop {\lim }\limits_{x \to – \infty } \dfrac{{x + 1}}{{\left| x \right|.\sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} – 3x}}\\ = \mathop {\lim }\limits_{x \to – \infty } \dfrac{{x + 1}}{{ – x\sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} – 3x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \to – \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = – x} \right)\\ = \mathop {\lim }\limits_{x \to – \infty } \dfrac{{1 + \frac{1}{x}}}{{ – \sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} – 3}}\\ = \dfrac{1}{{ – \sqrt 9 – 3}} = – \dfrac{1}{6}\end{array}\) Bình luận
$\lim\limits_{x\to -\infty}(\sqrt{9x^4+x+1}+3x)$ $=\lim\limits_{x\to -\infty} (x^2\sqrt{9+\dfrac{1}{x^3}+\dfrac{1}{x^4}}+3x)$ $=\lim\limits_{x\to -\infty} x^2\Big( \sqrt{9+\dfrac{1}{x^3}+\dfrac{1}{x^4}}+\dfrac{3}{x}\Big)$ $=+\infty$ Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {9{x^2} + x + 1} + 3x} \right)\\
= \mathop {\lim }\limits_{x \to – \infty } \dfrac{{\left( {\sqrt {9{x^2} + x + 1} + 3x} \right)\left( {\sqrt {9{x^2} + x + 1} – 3x} \right)}}{{\sqrt {9{x^2} + x + 1} – 3x}}\\
= \mathop {\lim }\limits_{x \to – \infty } \dfrac{{\left( {9{x^2} + x + 1} \right) – {{\left( {3x} \right)}^2}}}{{\sqrt {{x^2}\left( {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} \right)} – 3x}}\\
= \mathop {\lim }\limits_{x \to – \infty } \dfrac{{x + 1}}{{\left| x \right|.\sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} – 3x}}\\
= \mathop {\lim }\limits_{x \to – \infty } \dfrac{{x + 1}}{{ – x\sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} – 3x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \to – \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = – x} \right)\\
= \mathop {\lim }\limits_{x \to – \infty } \dfrac{{1 + \frac{1}{x}}}{{ – \sqrt {9 + \frac{1}{x} + \frac{1}{{{x^2}}}} – 3}}\\
= \dfrac{1}{{ – \sqrt 9 – 3}} = – \dfrac{1}{6}
\end{array}\)
$\lim\limits_{x\to -\infty}(\sqrt{9x^4+x+1}+3x)$
$=\lim\limits_{x\to -\infty} (x^2\sqrt{9+\dfrac{1}{x^3}+\dfrac{1}{x^4}}+3x)$
$=\lim\limits_{x\to -\infty} x^2\Big( \sqrt{9+\dfrac{1}{x^3}+\dfrac{1}{x^4}}+\dfrac{3}{x}\Big)$
$=+\infty$