Câu 1:
a. `lim(n-2)/(\sqrt[3n^2+1])=lim\frac{n(1-2/n)}{n\sqrt[3+1/n^2]}=(1-0)/\sqrt[3+0]=1/\sqrt[3]`
b. `\lim_{x \to1} \frac{-x^2+5x-4}{x^3-1}=\lim_{x \to1} \frac{-(x-4)(x-1)}{(x-1)(x^2+x+1)}=\lim_{x \to1} \frac{-x+4}{x^2+x+1}=1`
Câu 2:
`\lim_{x \to3^+} f(x)=\lim_{x \to3^+}\frac{2-\sqrt[x+1]}{3-x}=\lim_{x \to3^+} \frac{3-x}{(3-x)(2+\sqrt[x+1])}=\lim_{x \to3^+} \frac{1}{2+\sqrt[x+1]}=1/4`
`\lim_{x \to3^-} f(x)=f(3)=3/2a`
Để hàm số liên tục tại `x_0=3` thì:
`⇔ 1/4=3/2a`
`⇔ a=1/6`
