Đáp án:
$12)TCĐ: x=\pm 2; TCN: y=1\\ 13)TCN:y= \pm 1\\ 14)TCĐ: x=-2\\ 15)TCN:y= 0$
Giải thích các bước giải:
$12) y=\dfrac{x^2-x-2017}{\sqrt{x^4-16}} \\ ĐKXĐ:x^4-16>0\\ \Leftrightarrow (x-2)(x+2)(x^2+4)>0\\ \Leftrightarrow \left\{\begin{array}{l} x>2\\ x<-2\end{array} \right.\\ \displaystyle\lim_{x \to-2} \dfrac{x^2-x-2017}{\sqrt{x^4-16}}\\ =\displaystyle\lim_{x \to-2} \dfrac{-2011}{\sqrt{x^4-16}}\\ =- \infty\\ \Rightarrow TCĐ: x=-2\\ \displaystyle\lim_{x \to 2} \dfrac{x^2-x-2017}{\sqrt{x^4-16}}\\ =\displaystyle\lim_{x \to 2} \dfrac{-2015}{\sqrt{x^4-16}}\\ =- \infty\\ \Rightarrow TCĐ: x=2\\ \displaystyle\lim_{x \to \pm \infty} \dfrac{x^2-x-2017}{\sqrt{x^4-16}} \\ =\displaystyle\lim_{x \to \pm \infty} \dfrac{x^2-x-2017}{x^2\sqrt{1-\dfrac{16}{x^4}}} =1\\ \Rightarrow TCN: y=1\\ 13)y=\dfrac{\sqrt{x^2-x-2}}{x} \\ ĐKXĐ: \left\{\begin{array}{l} x^2-x-2\\ x \ne 0\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} x^2-x-2 \ge 0\\ x \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} x\ge 2\\ x \le -1\end{array} \right.\\ x \ne 0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x\ge 2\\ x \le -1\end{array} \right.$
Hàm số không có $TCĐ$
$\displaystyle\lim_{x \to -\infty} \dfrac{\sqrt{x^2-x-2}}{x} \\ =\displaystyle\lim_{x \to -\infty} \dfrac{|x|\sqrt{1-\dfrac{1}{x}-\dfrac{2}{x^2}}}{x} \\ =\displaystyle\lim_{x \to -\infty} \dfrac{-x}{x} \\ =-1\\ \Rightarrow TCN:y= -1\\ \displaystyle\lim_{x \to +\infty} \dfrac{\sqrt{x^2-x-2}}{x} \\ =\displaystyle\lim_{x \to +\infty} \dfrac{|x|\sqrt{1-\dfrac{1}{x}-\dfrac{2}{x^2}}}{x} \\ =\displaystyle\lim_{x \to +\infty} \dfrac{x}{x} \\ =1\\ \Rightarrow TCN:y= 1\\ c)y=\dfrac{\sqrt{4-x^2}}{x^2+5x+6}\\ ĐKXĐ: \left\{\begin{array}{l} 4-x^2 \ge 0 \\ x^2+5x+6 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} -2 \le x \le 2 \\ x\ne -2 ; x \ne -3 \end{array} \right.\\ \Leftrightarrow -2 < x \le 2 \\ y=\dfrac{\sqrt{4-x^2}}{x^2+5x+6}\\ =\dfrac{\sqrt{(2-x)(2+x)}}{(x+2)(x+3)}\\ \displaystyle\lim_{x \to -2} y\\ =\displaystyle\lim_{x \to -2} \dfrac{\sqrt{(2-x)(2+x)}}{(x+2)(x+3)}\\ =\displaystyle\lim_{x \to -2} \dfrac{\sqrt{2-x}}{\sqrt{x+2}(x+3)}\\ =\displaystyle\lim_{x \to -2} \dfrac{2}{\sqrt{x+2}(x+3)}\\ =\infty\\ \Rightarrow TCĐ: x=-2$
Hàm số không có $TCN$
$d)y=\dfrac{\sqrt{x-7}}{x^2+3x-4}\\ ĐKXĐ: \left\{\begin{array}{l} x-7 \ge 0 \\ x^2+3x-4 \ne 0\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} x \ge 7 \\ x \ne 1; x\ne -4 \end{array} \right.\\ \Leftrightarrow x \ge 7 $
Hàm số không có $TCĐ $
$\displaystyle\lim_{x \to +\infty} y\\ =\displaystyle\lim_{x \to +\infty} \dfrac{\sqrt{x-7}}{x^2+3x-4}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{\sqrt{\dfrac{1}{x^3}-\dfrac{7}{x^4}}}{1+\dfrac{3}{x}-\dfrac{4}{x^2}}\\ =0\\ \Rightarrow TCN:y= 0$
