giúp em vs ạ
Câu 3:
\(\lim_{x\to1^{+}}f\left(x\right)=\lim_{x\to1^{+}}\frac{\sqrt{4x-3}-x}{x^2-1}=\lim_{x\to1^{+}}\frac{4x-3-x^2}{\left(x^2-1\right)\cdot\left(\sqrt{4x-3}+x\right)}\)
\(=\lim_{x\to1^{+}}\frac{-\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x+1\right)\left(x+\sqrt{4x-3}\right)}=\lim_{x\to1^{+}}\frac{-\left(x-3\right)}{\left(x+1\right)\left(x+\sqrt{4x-3}\right)}\)
\(=\frac{-\left(1-3\right)}{\left(1+1\right)\left(1+\sqrt{4\cdot1-3}\right)}=\frac{3-1}{2\cdot\left(1+1\right)}=\frac{2}{2\cdot2}=\frac12\)
\(\lim_{x\to1^{-}}f\left(x\right)=\lim_{x\to1^{-}}2a-3\)
f(1)=2a-3
Để hàm số liên tục tại x=1 thì \(f\left(1\right)=\lim_{x\to1}f\left(x\right)\)
=>2a-3=1/2
=>2a=7/2
=>\(a=\frac74\)
Câu 2:
a: \(\lim_{x\to\left(-2\right)^{+}}3x+2=3\cdot\left(-2\right)+2=-6+2=-4<0\)
\(x^2-4=\left(-2\right)^2-4=0\)
=>\(\lim_{x\to\left(-2\right)^{+}}\frac{3x+2}{x^2-4}=-\) ∞
b: \(\lim_{x\to2}\frac{x^2-3x+2}{\sqrt{2x}-x}=\lim_{x\to2}\frac{\left(x-1\right)\left(x-2\right)}{\frac{2x-x^2}{\sqrt{2x}+x}}\)
\(=\lim_{x\to2}\frac{\left(x-1\right)\left(x-2\right)}{-x\left(x-2\right)}\cdot\left(x+\sqrt{2x}\right)=\lim_{x\to2}\frac{-\left(x-1\right)}{x}\cdot\left(x+\sqrt{2x}\right)=-\frac{\left(2-1\right)}{2}\cdot\left(2+\sqrt{2\cdot2}\right)=-\frac12\cdot\left(2+2\right)=-\frac12\cdot4=-2\)
