Đáp án:
\(a)\; \lim_{x\to +\;\infty}\dfrac{(2-x)^3}{2x^3+4x^2+1}=-\dfrac 12\)
\(c)\; \lim_{x\to -\; \infty} \Big(\sqrt{x^2-3x+5}-\sqrt{x^2+3}\Big)=\dfrac 32\)
Giải thích các bước giải:
\(a)\)
\(\lim_{x\to +\infty}\dfrac{(2-x)^3}{2x^3+4x^2+1}\)
\(=\lim_{x\to +\infty}\dfrac{\bigg(\dfrac 2x-1\bigg)^3}{2+\dfrac 4x+\dfrac 1{x^3}}\)
\(=\dfrac{(-1)^3}{2}\)
\(=-\dfrac 12\)
\(c)\)
\(\lim_{x\to -\; \infty} \Big(\sqrt{x^2-3x+5}-\sqrt{x^2+3}\Big)\)
\(=\lim_{x\to -\; \infty}\dfrac{-3x+2}{\sqrt{x^2-3x+5}+\sqrt{x^2+3}}\)
\(\lim_{x\to -\; \infty} \dfrac{x.\bigg(-3+\dfrac 2x\bigg)}{|x|.\Bigg(\sqrt{1-\dfrac 3x+\dfrac 5{x^2}} + \sqrt{1+\dfrac 3{x^2}}\Bigg)}\)
\(\lim_{x\to -\; \infty} \dfrac{x\bigg(-3+\dfrac 2x\bigg)}{-x\Bigg(\sqrt{1-\dfrac 3x+\dfrac 5{x^2}}+\sqrt{1+\dfrac 3{x^2}}\Bigg)}\)
\(=\dfrac{-3}{-\Big(\sqrt{1}+\sqrt{1}\Big)}\)
\(=\dfrac 32\)
