Câu 2:
\(\lim_{x\to3^{+}}f\left(x\right)=\lim_{x\to3^{+}}\frac{\sqrt{x-2}-1}{x-3}\)
\(=\lim_{x\to3^{+}}\frac{x-2-1}{\left(x-3\right)\left(\sqrt{x-2}+1\right)}=\lim_{x\to3^{+}}\frac{1}{\sqrt{x-2}+1}\)
\(=\frac{1}{\sqrt{3-2}+1}=\frac{1}{1+1}=\frac12\)
\(\lim_{x\to3^{-}}f\left(x\right)=\lim_{x\to3^{-}}2x^2-1=2\cdot3^2-1=2\cdot9-1=18-1=17\)
Vì \(17<>\frac12\)
nên không tồn tại \(\lim_{x\to3}f\left(x\right)\)
Câu 3:
\(\lim_{x\to1^{-}}f\left(x\right)=\lim_{x\to1^{-}}\frac{x^3-1}{-2x^2+5x-3}\)
\(=\lim_{x\to1^{-}}\frac{\left(x-1\right)\left(x^2+x+1\right)}{-\left(x-1\right)\left(2x-3\right)}=\lim_{x\to1^{-}}\frac{x^2+x+1}{-2x+3}\)
\(=\frac{1^2+1+1}{-2\cdot1+3}=\frac31=3\)
\(\lim_{x\to1^{+}}f\left(x\right)=\lim_{x\to1^{+}}2mx^2+3=2m\cdot1^2+3=2m+3\)
Để \(\lim_{x\to1}f\left(x\right)\) tồn tại thì 2m+3=3
=>2m=0
=>m=0
=>\(\lim_{x\to1}f\left(x\right)=3\)
