b: \(\lim_{x\to1}f\left(x\right)=\lim_{x\to1}\frac{x^3-3x+2}{\left(x-1\right)^2}\)
\(=\lim_{x\to1}\frac{x^3-x-2x+2}{\left(x-1\right)^2}\)
\(=\lim_{x\to1}\frac{x\left(x^2-1\right)-2\left(x-1\right)}{\left(x-1\right)^2}\)
\(=\lim_{x\to1}\frac{\left(x-1\right)\left(x^2+x-2\right)}{\left(x-1\right)^2}=\lim_{x\to1}\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)}=\lim_{x\to1}x+2\)
=1+2=3
\(f\left(1\right)=\sqrt{3\cdot1+1}+1=\sqrt4+1=2+1=3\)
=>\(\lim_{x\to1}f\left(x\right)=f\left(1\right)\)
=>f(x) liên tục khi x=1
=>f(x) liên tục trên R
c: \(\lim_{x\to1}f\left(x\right)=\lim_{x\to1}\frac{x-\sqrt{4x-3}}{x-1}\)
\(=\lim_{x\to1}\frac{x^2-4x+3}{\left(x-1\right)\left(x+\sqrt{4x-3}\right)}=\lim_{x\to1}\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x+\sqrt{4x-3}\right)}\)
\(=\lim_{x\to1}\frac{x-3}{x+\sqrt{4x-3}}=\frac{1-3}{1+\sqrt{4\cdot1-3}}=\frac{-2}{1+\sqrt{4-3}}=\frac{-2}{1+1}=-1\)
\(f\left(1\right)=3\cdot1-5=3-5=-2\)
=>\(\lim_{x\to1}f\left(x\right)<>f\left(1\right)\)
=>f(x) gián đoạn tại x=1
d: \(\lim_{x\to0}f\left(x\right)=\lim_{x\to0}\frac{\sqrt{3x^2+1}-1}{x^2}\)
\(=\lim_{x\to0}\frac{3x^2+1-1}{x^2\left(\sqrt{3x^2+1}+1\right)}=\lim_{x\to0}\frac{3x^2}{x^2\left(\sqrt{3x^2+1}+1\right)}\)
\(=\lim_{x\to0}\frac{3}{\sqrt{3x^2+1}+1}=\frac{3}{\sqrt{3+1}+1}=\frac{3}{2+1}=\frac33=1\)
\(f\left(0\right)=0+2=2\)
=>\(\lim_{x\to0}f\left(x\right)<>f\left(0\right)\)
=>Hàm số gián đoạn tại x=0
h: \(\lim_{x\to0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}\)
\(=\lim_{x\to0}\frac{\sqrt{x+1}-1+\sqrt{x+4}-2}{x}=\lim_{x\to0}\frac{\frac{x+1-1}{\sqrt{x+1}+1}+\frac{x+4-4}{\sqrt{x+4}+2}}{x}\)
\(=\lim_{x\to0}\frac{1}{\sqrt{x+1}+1}+\frac{1}{\sqrt{x+4}+2}=\frac{1}{\sqrt{0+1}+1}+\frac{1}{\sqrt{0+4}+2}\)
\(=\frac12+\frac14=\frac34\)
