$\lim_{x \to -\infty} $ ($\sqrt[]{x^{2}+4}$+x ) Lam ho voi ah 30/10/2021 Bởi Abigail $\lim_{x \to -\infty} $ ($\sqrt[]{x^{2}+4}$+x ) Lam ho voi ah
Đáp án: 0 Giải thích các bước giải: \(\begin{array}{l}\mathop {\lim }\limits_{x \to – \infty } \dfrac{{{x^2} + 4 – {x^2}}}{{\sqrt {{x^2} + 4} – x}}\\ = \mathop {\lim }\limits_{x \to – \infty } \dfrac{4}{{\sqrt {{x^2} + 4} – x}}\\ = \mathop {\lim }\limits_{x \to – \infty } \dfrac{{\dfrac{4}{x}}}{{ – \sqrt {1 + \dfrac{4}{{{x^2}}}} – 1}} = \dfrac{0}{{ – 2}} = 0\end{array}\) Bình luận
$\lim\limits_{x\to -\infty}(\sqrt{x^2+4}+x)$ $=\lim\limits_{x\to -\infty}\dfrac{x^2+4-x^2}{\sqrt{x^2+4}-x}$ $=\lim\limits_{x\to -\infty}\dfrac{4}{\sqrt{x^2+4}-x}$ $=\lim\limits_{x\to -\infty}\dfrac{4}{-x\sqrt{1+\dfrac{4}{x^2}}-x}$ $=\lim\limits_{x\to -\infty}\dfrac{ \dfrac{-4}{x}}{\sqrt{1+\dfrac{4}{x^2}}+1}$ $=0$ Bình luận
Đáp án:
0
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to – \infty } \dfrac{{{x^2} + 4 – {x^2}}}{{\sqrt {{x^2} + 4} – x}}\\
= \mathop {\lim }\limits_{x \to – \infty } \dfrac{4}{{\sqrt {{x^2} + 4} – x}}\\
= \mathop {\lim }\limits_{x \to – \infty } \dfrac{{\dfrac{4}{x}}}{{ – \sqrt {1 + \dfrac{4}{{{x^2}}}} – 1}} = \dfrac{0}{{ – 2}} = 0
\end{array}\)
$\lim\limits_{x\to -\infty}(\sqrt{x^2+4}+x)$
$=\lim\limits_{x\to -\infty}\dfrac{x^2+4-x^2}{\sqrt{x^2+4}-x}$
$=\lim\limits_{x\to -\infty}\dfrac{4}{\sqrt{x^2+4}-x}$
$=\lim\limits_{x\to -\infty}\dfrac{4}{-x\sqrt{1+\dfrac{4}{x^2}}-x}$
$=\lim\limits_{x\to -\infty}\dfrac{ \dfrac{-4}{x}}{\sqrt{1+\dfrac{4}{x^2}}+1}$
$=0$