Question

tính các giới hạn sau:

a/lim\(\frac{\sqrt{3x+1}-\sqrt{x+3}}{x^2-1}\) (x-->1)

b/lim\(\frac{x^3+x+1}{x-1}\) (x-->1⁺)

c/lim\(\frac{x-15}{x-2}\) (x-->2⁺)

giúp mình với ạ

Answer 1

\(a=\lim\limits_{x\rightarrow1}\frac{\left(\sqrt{3x+1}-\sqrt{x+3}\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}=\lim\limits_{x\rightarrow1}\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{2}{\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}=\frac{2}{2.4}=\frac{1}{4}\)

\(b=\frac{3}{0}=+\infty\)

\(c=\frac{-13}{0}=-\infty\)