Đáp án:
$5.TCĐ: x=0\\ 6.TCĐ: x=\pm 3; TCN: y=1\\ 7.TCĐ: x=\pm 1; TCN: y=0\\ 8.TCN: y=1$
Giải thích các bước giải:
$5. y=\dfrac{\sqrt{-4x^2+3x+1}}{2x}\\ ĐKXĐ: \left\{\begin{array}{l} -4x^2+3x+1 \ge 0 \\ 2x \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} -\dfrac{1}{4} \le x \le 1 \\ x \ne 0\end{array} \right.$ Hàm số không có $TCN$
$\displaystyle\lim_{x \to 0} y\\ =\displaystyle\lim_{x \to 0} \dfrac{1}{2x}\\ =\infty\\ \Rightarrow TCĐ: x=0\\ 6. y=\dfrac{\sqrt{x^2+x+2}}{|x|-3}\\ ĐKXĐ: \left\{\begin{array}{l} x^2+x+2 \ge 0 \\ |x|-3 \ne 0\end{array} \right.\\ \Leftrightarrow x \ne \pm 3\\ \displaystyle\lim_{x \to 3} y\\ =\displaystyle\lim_{x \to 3} \dfrac{\sqrt{x^2+x+2}}{|x|-3}\\ =\displaystyle\lim_{x \to 3} \dfrac{14}{|x|-3}\\ =\infty\\ \Rightarrow TCĐ: x=-3\\ \displaystyle\lim_{x \to -3} y\\ =\displaystyle\lim_{x \to -3} \dfrac{\sqrt{x^2+x+2}}{|x|-3}\\ =\displaystyle\lim_{x \to -3} \dfrac{8}{|x|-3}\\ =\infty\\ \Rightarrow TCĐ: x=3\\ \displaystyle\lim_{x \to \pm \infty} y\\ =\displaystyle\lim_{x \to \pm \infty} \dfrac{\sqrt{x^2+x+2}}{|x|-3}\\ =\displaystyle\lim_{x \to \pm \infty} \dfrac{|x|\sqrt{1+\dfrac{1}{x}+\dfrac{2}{x}}}{|x|-3}\\ =1\\ \Rightarrow TCN: y=1\\ 7. y=\dfrac{\sqrt{x+1}}{x^2-1}\\ ĐKXĐ: \left\{\begin{array}{l} x+1 \ge 0 \\ x^2-1 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge -1 \\ x\ne \pm 1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > -1 \\ x\ne 1\end{array} \right.\\ \displaystyle\lim_{x \to -1} y\\ =\displaystyle\lim_{x \to -1} \dfrac{\sqrt{x+1}}{x^2-1}\\ =\displaystyle\lim_{x \to -1} \dfrac{\sqrt{x+1}}{(x-1)(x+1)}\\ =\displaystyle\lim_{x \to -1} \dfrac{1}{(x-1)\sqrt{x+1}}\\ =-\infty\\ \Rightarrow TCĐ: x=-1\\ \displaystyle\lim_{x \to 1} y\\ =\displaystyle\lim_{x \to 1} \dfrac{\sqrt{x+1}}{x^2-1}\\ =\displaystyle\lim_{x \to 1} \dfrac{\sqrt{x+1}}{(x-1)(x+1)}\\ =\displaystyle\lim_{x \to 1} \dfrac{1}{(x-1)\sqrt{x+1}}\\ =\infty\\ \Rightarrow TCĐ: x=1\\ \displaystyle\lim_{x \to + \infty} y\\ =\displaystyle\lim_{x \to +\infty} \dfrac{\sqrt{x+1}}{x^2-1}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{\sqrt{\dfrac{1}{x^3}+\dfrac{1}{x^4}}}{1-\dfrac{1}{x^2}}\\ =0\\ \Rightarrow TCN: y=0\\ 8)y=\sqrt{x^2+2x+3}-x \ \ \ \ D=\mathbb{R}$
Hàm số không có $TCĐ$
$\displaystyle\lim_{x \to +\infty} y\\ =\displaystyle\lim_{x \to +\infty} \sqrt{x^2+2x+3}-x\\ =\displaystyle\lim_{x \to +\infty} \dfrac{(\sqrt{x^2+2x+3}-x)(\sqrt{x^2+2x+3}+x)}{\sqrt{x^2+2x+3}+x}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{2x+3}{\sqrt{x^2+2x+3}+x}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{2+\dfrac{3}{x}}{\sqrt{1+\dfrac{2}{x}+\dfrac{3}{x^2}}+1}\\ =1\\ \Rightarrow TCN: y=1$
