Tinh gioi han hàm số lim x->- vô cùng căn 4x²-x +2x 20/11/2021 Bởi Lyla Tinh gioi han hàm số lim x->- vô cùng căn 4x²-x +2x
$=\lim\limits_{x\to -\infty}\dfrac{4x^2-x-4x^2}{\sqrt{4x^2-x}-2x}$ $=\lim\limits_{x\to -\infty}\dfrac{-x}{\sqrt{x^2(4-\dfrac{1}{x})}-2x}$ $=\lim\limits_{x\to -\infty}\dfrac{-1}{-\sqrt{4-\dfrac{1}{x}}-2}$ $=\lim\limits_{x\to -\infty}\dfrac{1}{\sqrt{4-\dfrac{1}{x}}+2}$ $=\dfrac{1}{4}$ Bình luận
Đáp án: \[\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {4{x^2} – x} + 2x} \right) = \frac{1}{4}\] Giải thích các bước giải: Ta có: \(\begin{array}{l}\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {4{x^2} – x} + 2x} \right)\\ = \mathop {\lim }\limits_{x \to – \infty } \frac{{\left( {\sqrt {4{x^2} – x} + 2x} \right)\left( {\sqrt {4{x^2} – x} – 2x} \right)}}{{\sqrt {4{x^2} – x} – 2x}}\\ = \mathop {\lim }\limits_{x \to – \infty } \frac{{\left( {4{x^2} – x} \right) – {{\left( {2x} \right)}^2}}}{{\sqrt {{x^2}\left( {4 – \frac{1}{x}} \right)} – 2x}}\\ = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – x}}{{\left| x \right|.\sqrt {4 – \frac{1}{x}} – 2x}}\\ = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – x}}{{ – x\sqrt {4 – \frac{1}{x}} – 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \to – \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = – x} \right)\\ = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 1}}{{ – \sqrt {4 – \frac{1}{x}} – 2}}\\ = \frac{{ – 1}}{{ – \sqrt 4 – 2}} = \frac{1}{4}\end{array}\) Vậy \(\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {4{x^2} – x} + 2x} \right) = \frac{1}{4}\) Bình luận
$=\lim\limits_{x\to -\infty}\dfrac{4x^2-x-4x^2}{\sqrt{4x^2-x}-2x}$
$=\lim\limits_{x\to -\infty}\dfrac{-x}{\sqrt{x^2(4-\dfrac{1}{x})}-2x}$
$=\lim\limits_{x\to -\infty}\dfrac{-1}{-\sqrt{4-\dfrac{1}{x}}-2}$
$=\lim\limits_{x\to -\infty}\dfrac{1}{\sqrt{4-\dfrac{1}{x}}+2}$
$=\dfrac{1}{4}$
Đáp án:
\[\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {4{x^2} – x} + 2x} \right) = \frac{1}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {4{x^2} – x} + 2x} \right)\\
= \mathop {\lim }\limits_{x \to – \infty } \frac{{\left( {\sqrt {4{x^2} – x} + 2x} \right)\left( {\sqrt {4{x^2} – x} – 2x} \right)}}{{\sqrt {4{x^2} – x} – 2x}}\\
= \mathop {\lim }\limits_{x \to – \infty } \frac{{\left( {4{x^2} – x} \right) – {{\left( {2x} \right)}^2}}}{{\sqrt {{x^2}\left( {4 – \frac{1}{x}} \right)} – 2x}}\\
= \mathop {\lim }\limits_{x \to – \infty } \frac{{ – x}}{{\left| x \right|.\sqrt {4 – \frac{1}{x}} – 2x}}\\
= \mathop {\lim }\limits_{x \to – \infty } \frac{{ – x}}{{ – x\sqrt {4 – \frac{1}{x}} – 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \to – \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = – x} \right)\\
= \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 1}}{{ – \sqrt {4 – \frac{1}{x}} – 2}}\\
= \frac{{ – 1}}{{ – \sqrt 4 – 2}} = \frac{1}{4}
\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {4{x^2} – x} + 2x} \right) = \frac{1}{4}\)