Đáp án:
$a) y=\pm \dfrac{1}{5}\\ b)y=-3.$
Giải thích các bước giải:
$a)y=\dfrac{\sqrt{x^2+4}}{5x+1} \ \ \ \ D=\mathbb{R} \setminus \left\{-\dfrac{1}{5}\right\}\\ =\dfrac{|x|\sqrt{1+\dfrac{4}{x^2}}}{5x+1}\\ \displaystyle \lim_{x \to +\infty} \dfrac{|x|\sqrt{1+\dfrac{4}{x^2}}}{5x+1}=\displaystyle \lim_{x \to +\infty} \dfrac{x\sqrt{1+\dfrac{4}{x^2}}}{5x+1}=\displaystyle \lim_{x \to +\infty} \dfrac{\sqrt{1+\dfrac{4}{x^2}}}{5+\dfrac{1}{x}}=\dfrac{1}{5}\\ \displaystyle \lim_{x \to -\infty} \dfrac{|x|\sqrt{1+\dfrac{4}{x^2}}}{5x+1}=\displaystyle \lim_{x \to -\infty} \dfrac{-x\sqrt{1+\dfrac{4}{x^2}}}{5x+1}=\displaystyle \lim_{x \to +\infty} \dfrac{-\sqrt{1+\dfrac{4}{x^2}}}{5+\dfrac{1}{x}}=-\dfrac{1}{5}\\ \Rightarrow \text{TCN: } y=\pm \dfrac{1}{5}\\ b)y=-3+\dfrac{1}{\sqrt{x}} \ \ \ \ D=(0;+\infty)\\ =\dfrac{1-3\sqrt{x}}{\sqrt{x}}\\ \displaystyle \lim_{x \to +\infty} \dfrac{1-3\sqrt{x}}{\sqrt{x}}\\=\displaystyle \lim_{x \to +\infty} \dfrac{\dfrac{1}{\sqrt{x}}-3}{1}=-3 \\\Rightarrow \text{TCN: } y=-3.$
