lim x tiến tới âm vô cực căn(4x^2-8x+10)+2x

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1. Ta có

   $\underset{x \to -\infty}{\lim} (\sqrt{4x^2 – 8x + 10} + 2x) = \underset{x \to -\infty}{\lim} \dfrac{4x^2 -8x + 10 – 4x^2}{\sqrt{4x^2 – 8x + 10} – 2x}$

   $= \underset{x \to -\infty}{\lim} \dfrac{10 – 8x}{\sqrt{4x^2 – 8x + 10} – 2x}$

   $= \underset{x \to +\infty}{\lim}\dfrac{-\frac{10}{x} + 8}{\sqrt{4 – \frac{8}{x} + \frac{10}{x^2}} + 2}$

   $= \dfrac{8}{2 + 2} = 2$

   Vậy

   $\underset{x \to -\infty}{\lim} (\sqrt{4x^2 – 8x + 10} + 2x) = 2$.

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2. Lời giải:

   $\lim\limits_{x \to -\infty} (\sqrt[]{4x^2-8x+10}+2x)$

   $=\lim\limits_{x \to -\infty} \dfrac{(\sqrt[]{4x^2-8x+10}+2x)(\sqrt[]{4x^2-8x+10}-2x)}{\sqrt[]{4x^2-8x+10}-2x}$

   $=\lim\limits_{x \to -\infty} \dfrac{(4x^2-8x+10)-(2x)^2}{\sqrt[]{4x^2-8x+10}-2x}$

   $=\lim\limits_{x \to -\infty} \dfrac{-8x+10}{\sqrt[]{4x^2-8x+10}-2x}$

   $=\lim\limits_{x \to -\infty} \dfrac{-8+\dfrac{10}{x}}{-\sqrt[]{4-\dfrac{8}{x}+\dfrac{10}{x^2}}-2}$

   $=\dfrac{-8}{-4}=2$

   Giải thích:

   Do $\sqrt{A^2}=|A|>0$ $(\forall A)$

   $\to$ khi $x\to-\infty$ thì $x=-\sqrt{x^2}$

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