Mọi người ơi giúp mình với lim $_{±∞}$ 2x+$\sqrt[]{4x^{2}+1}$ 28/11/2021 Bởi Autumn Mọi người ơi giúp mình với lim $_{±∞}$ 2x+$\sqrt[]{4x^{2}+1}$
Giải thích các bước giải: Ta có: \(\begin{array}{l}*)\\\mathop {\lim }\limits_{x \to + \infty } \left( {2x + \sqrt {4{x^2} + 1} } \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \left( {2x + \sqrt {{x^2}\left( {4 + \frac{1}{{{x^2}}}} \right)} } \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \left( {2x + x.\sqrt {4 + \frac{1}{{{x^2}}}} } \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {2 + \sqrt {4 + \frac{1}{{{x^2}}}} } \right)} \right]\\\mathop {\lim }\limits_{x \to + \infty } x = + \infty \\\mathop {\lim }\limits_{x \to + \infty } \left( {2 + \sqrt {4 + \frac{1}{{{x^2}}}} } \right) = 2 + \sqrt {4 + 0} = 4\\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {2 + \sqrt {4 + \frac{1}{{{x^2}}}} } \right)} \right] = + \infty \\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {2x + \sqrt {4{x^2} + 1} } \right) = + \infty \\*)\\\mathop {\lim }\limits_{x \to – \infty } \left( {2x + \sqrt {4{x^2} + 1} } \right)\\ = \mathop {\lim }\limits_{x \to – \infty } \frac{{\left( {2x + \sqrt {4{x^2} + 1} } \right)\left( {2x – \sqrt {4{x^2} + 1} } \right)}}{{2x – \sqrt {4{x^2} + 1} }}\\ = \mathop {\lim }\limits_{x \to – \infty } \frac{{{{\left( {2x} \right)}^2} – \left( {4{x^2} + 1} \right)}}{{2x – \sqrt {4{x^2} + 1} }}\\ = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 1}}{{2x – \sqrt {4{x^2} + 1} }}\\\mathop {\lim }\limits_{x \to – \infty } \left( {2x – \sqrt {4{x^2} + 1} } \right) = \mathop {\lim }\limits_{x \to – \infty } \left( {2x – \sqrt {{x^2}\left( {4 + \frac{1}{{{x^2}}}} \right)} } \right)\\ = \mathop {\lim }\limits_{x \to – \infty } \left( {2x – \left| x \right|.\sqrt {4 + \frac{1}{{{x^2}}}} } \right) = \mathop {\lim }\limits_{x \to – \infty } \left( {2x + x.\sqrt {4 + \frac{1}{{{x^2}}}} } \right)\\\left( {x \to – \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = – x} \right)\\ = \mathop {\lim }\limits_{x \to – \infty } \left[ {x.\left( {2 + \sqrt {4 + \frac{1}{{{x^2}}}} } \right)} \right] = \left( { – \infty } \right).\left( {2 + \sqrt 4 } \right) = – \infty \\ \Rightarrow \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 1}}{{2x – \sqrt {4{x^2} + 1} }} = 0\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
\mathop {\lim }\limits_{x \to + \infty } \left( {2x + \sqrt {4{x^2} + 1} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {2x + \sqrt {{x^2}\left( {4 + \frac{1}{{{x^2}}}} \right)} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {2x + x.\sqrt {4 + \frac{1}{{{x^2}}}} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {2 + \sqrt {4 + \frac{1}{{{x^2}}}} } \right)} \right]\\
\mathop {\lim }\limits_{x \to + \infty } x = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \left( {2 + \sqrt {4 + \frac{1}{{{x^2}}}} } \right) = 2 + \sqrt {4 + 0} = 4\\
\Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {2 + \sqrt {4 + \frac{1}{{{x^2}}}} } \right)} \right] = + \infty \\
\Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {2x + \sqrt {4{x^2} + 1} } \right) = + \infty \\
*)\\
\mathop {\lim }\limits_{x \to – \infty } \left( {2x + \sqrt {4{x^2} + 1} } \right)\\
= \mathop {\lim }\limits_{x \to – \infty } \frac{{\left( {2x + \sqrt {4{x^2} + 1} } \right)\left( {2x – \sqrt {4{x^2} + 1} } \right)}}{{2x – \sqrt {4{x^2} + 1} }}\\
= \mathop {\lim }\limits_{x \to – \infty } \frac{{{{\left( {2x} \right)}^2} – \left( {4{x^2} + 1} \right)}}{{2x – \sqrt {4{x^2} + 1} }}\\
= \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 1}}{{2x – \sqrt {4{x^2} + 1} }}\\
\mathop {\lim }\limits_{x \to – \infty } \left( {2x – \sqrt {4{x^2} + 1} } \right) = \mathop {\lim }\limits_{x \to – \infty } \left( {2x – \sqrt {{x^2}\left( {4 + \frac{1}{{{x^2}}}} \right)} } \right)\\
= \mathop {\lim }\limits_{x \to – \infty } \left( {2x – \left| x \right|.\sqrt {4 + \frac{1}{{{x^2}}}} } \right) = \mathop {\lim }\limits_{x \to – \infty } \left( {2x + x.\sqrt {4 + \frac{1}{{{x^2}}}} } \right)\\
\left( {x \to – \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = – x} \right)\\
= \mathop {\lim }\limits_{x \to – \infty } \left[ {x.\left( {2 + \sqrt {4 + \frac{1}{{{x^2}}}} } \right)} \right] = \left( { – \infty } \right).\left( {2 + \sqrt 4 } \right) = – \infty \\
\Rightarrow \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 1}}{{2x – \sqrt {4{x^2} + 1} }} = 0
\end{array}\)