lim(căn x^2+3x-2 - căn x^2+5) X tiến đến + vô cực
2 câu trả lời
Đáp án:
$\dfrac{3}{2}.$
Giải thích các bước giải:
$\displaystyle\lim_{x \to +\infty} \left(\sqrt{x^2+3x-3}-\sqrt{x^2+5}\right)\\ =\displaystyle\lim_{x \to +\infty} \left(\sqrt{x^2+3x-3}-x-\sqrt{x^2+5}+x\right)\\ =\displaystyle\lim_{x \to +\infty} \left(\sqrt{x^2+3x-3}-x\right)+\displaystyle\lim_{x \to +\infty} \left(x-\sqrt{x^2+5}\right)\\ =\displaystyle\lim_{x \to +\infty} \dfrac{\left(\sqrt{x^2+3x-3}-x\right)\left(\sqrt{x^2+3x-3}+x\right)}{\sqrt{x^2+3x-3}+x}+\displaystyle\lim_{x \to +\infty} \dfrac{\left(x-\sqrt{x^2+5}\right) \left(x+\sqrt{x^2+5}\right)}{ x+\sqrt{x^2+5}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{x^2+3x-3-x^2}{\sqrt{x^2+3x-3}+x}+\displaystyle\lim_{x \to +\infty} \dfrac{x^2-(x^2+5)}{ x+\sqrt{x^2+5}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{3x-3}{\sqrt{x^2+3x-3}+x}+\displaystyle\lim_{x \to +\infty} \dfrac{-5}{ x+\sqrt{x^2+5}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{3-\dfrac{3}{x}}{\sqrt{1+\dfrac{3}{x}-\dfrac{3}{x^2}}+1}+\displaystyle\lim_{x \to +\infty} \dfrac{\dfrac{-5}{x}}{ 1+\sqrt{1+\dfrac{5}{x^2}}}\\ =\dfrac{3}{2}.$
