Đáp án:
$30)\text{TCN: }y=0$
Hàm số không có TCĐ
$31) \text{TCN: }y=\pm 1; \text{TCĐ: } x=\pm 10.$
Giải thích các bước giải:
$30)\\ y=\dfrac{x\sqrt{x}}{x^2+4x+5} \ \ \ \ D=[0;+\infty)\\ \displaystyle\lim_{x \to +\infty} \dfrac{x\sqrt{x}}{x^2+4x+5}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{\dfrac{1}{\sqrt{x}}}{1+\dfrac{4}{x}+\dfrac{5}{x^2}}\\ =0\\ \Rightarrow \text{TCN: }y=0$
Hàm số không có TCĐ do mẫu vô nghiệm
$31)\\ y=\dfrac{x}{\sqrt{x^2-100}} \ \ \ \ D=(-\infty;-10) \cup (10;+\infty)\\ \displaystyle\lim_{x \to +\infty} \dfrac{x}{\sqrt{x^2-100}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{x}{|x|\sqrt{1-\dfrac{100}{x^2}}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{x}{x\sqrt{1-\dfrac{100}{x^2}}}\\ =\displaystyle\lim_{x \to +\infty} \dfrac{1}{\sqrt{1-\dfrac{100}{x^2}}}\\ =1\\ \displaystyle\lim_{x \to -\infty} \dfrac{x}{\sqrt{x^2-100}}\\ =\displaystyle\lim_{x \to -\infty} \dfrac{x}{|x|\sqrt{1-\dfrac{100}{x^2}}}\\ =\displaystyle\lim_{x \to -\infty} \dfrac{x}{-x\sqrt{1-\dfrac{100}{x^2}}}\\ =\displaystyle\lim_{x \to -\infty} \dfrac{1}{-\sqrt{1-\dfrac{100}{x^2}}}\\ =-1\\ \Rightarrow \text{TCN: }y=\pm 1\\ \displaystyle\lim_{x \to 10^+} \dfrac{x}{\sqrt{x^2-100}} =+\infty\\ \displaystyle\lim_{x \to -10^+} \dfrac{x}{\sqrt{x^2-100}} =-\infty\\ \Rightarrow \text{TCĐ: } x=\pm 10.$
