\(\lim\limits_{x\rightarrow4}\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{2+2}{2+1}=\frac{4}{3}\)
\(\lim\limits_{x\rightarrow3}\frac{\left(\sqrt{x^2+16}-5\right)\left(\sqrt{x^2+16}+5\right)}{x\left(x-3\right)\left(\sqrt{x^2+16}+5\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(\sqrt{x^2+16}+5\right)}=\lim\limits_{x\rightarrow3}\frac{x+3}{x\left(\sqrt{x^2+16}+5\right)}=\frac{1}{5}\)
\(\lim\limits_{x\rightarrow0}\frac{2-3\sqrt{x}}{1+9\sqrt{x}}=\frac{2-3.0}{1+9.0}=2\) (ko phải dạng vô định, cứ thay số)
\(\lim\limits_{x\rightarrow1}\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x^2+x+1\right)}=\frac{1+2}{\left(1+1\right)\left(1+1+1\right)}=\frac{1}{2}\)
