\(\lim\limits_{x\rightarrow2^-}\left(\dfrac{1}{x-2}-\dfrac{1}{x^2-4}\right)\)

\(=\lim\limits_{x\rightarrow2^-}\dfrac{x+2-1}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2^-}\dfrac{x+1}{\left(x-2\right)\left(x+2\right)}\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2^-}\dfrac{x+1}{x+2}=\dfrac{2+1}{2+2}=\dfrac{3}{4}>0\\x-2< 0\end{matrix}\right.\)

Giới hạn đã cho hữu hạn nên \(ax^3+bx^2+4=0\) có nghiệm \(x=-2\)

\(\Rightarrow-8a+4b+4=0\Rightarrow b=2a-1\)

\(\lim\limits_{x\rightarrow-2}\dfrac{ax^3+\left(2a-1\right)x^2+4}{\left(x-1\right)^2\left(x+2\right)}=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(ax^2-x+2\right)}{\left(x-1\right)^2\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{ax^2-x+2}{\left(x-1\right)^2}=\dfrac{4a+4}{9}=2\Rightarrow a=\dfrac{7}{2}\) \(\Rightarrow b=6\)

\(\lim\limits_{x\rightarrow2}\dfrac{\left(3x-5\right)}{\left(x-2\right)^2}=+\infty\)

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}3x-5=3\cdot2-5=1>0\\\left(x-2\right)^2>0\\\lim\limits_{x\rightarrow2}\left(x-2\right)^2=\left(2-2\right)^2=0\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-2\right)\left(3x+2\right)}+\dfrac{1}{\left(x-2\right)\left(x-10\right)}\right)=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)}\left(\dfrac{x-10+3x+2}{\left(3x+2\right)\left(x-10\right)}\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(3x+2\right)\left(x-10\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\left(3x+2\right)\left(x-10\right)}=-\dfrac{1}{16}\)

\(=\frac{3\sqrt{3}}{0^+}=+\infty\)

a: \(\lim\limits_{x\rightarrow2}\dfrac{1-\sqrt{x^2+3}}{-x^2+3x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+3}-1}{x^2-3x+2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{2^2+3}-1}{2^2-3\cdot2+2}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\sqrt{2^2+3}-1=\sqrt{7}-1>0\\\lim\limits_{x\rightarrow2}2^2-3\cdot2+2=0\end{matrix}\right.\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{4x-1}+3}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4x-1-9}{\sqrt{4x-1}-3}\cdot\dfrac{1}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}\cdot\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}=\dfrac{4\cdot2-10}{\sqrt{4\cdot2-1}-3}=\dfrac{-2}{\sqrt{7}-3}>0\\\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{1}{\left(2+2\right)\cdot\left(2-2\right)}=+\infty\end{matrix}\right.\)

a: \(=lim_{x->2}\dfrac{x^3-2x^2+4x^2-8x+2x-4}{-\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=lim_{x->2}\dfrac{\left(x-2\right)\left(x^2+4x+2\right)}{-\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=lim_{x->2}\dfrac{-x^2-4x-2}{x^2+2x+4}\)

\(=lim_{x->2}\dfrac{-1-\dfrac{4}{x}-\dfrac{2}{x^2}}{1+\dfrac{2}{x}+\dfrac{4}{x^2}}=\dfrac{-1}{1}=-1\)

b: \(lim_{x->2}\dfrac{x^3-2x^2+3x^2-6x+x-2}{\left(x-2\right)\left(x-1\right)}\)

\(=lim_{x->2}\dfrac{\left(x-2\right)\left(x^2+3x+1\right)}{\left(x-2\right)\left(x-1\right)}\)

\(=lim_{x->2}\dfrac{x^2+3x+1}{x-1}\)

\(=lim_{x->2}\dfrac{1+\dfrac{3}{x}+\dfrac{1}{x^2}}{\dfrac{1}{x}-\dfrac{1}{x^2}}\)

lim(1+3/x+1/x^2)=1>0

lim(1/x-1/x^2)=(x-1)/x^2<0

=>lim=dương vô cực

\(=\lim\limits_{x\rightarrow2^-}\frac{-\left(x+2\right)\sqrt{\left(2-x\right)^2}}{\sqrt{\left(x^2+1\right)\left(2-x\right)}}=\lim\limits_{x\rightarrow2^-}\frac{-\left(x+2\right)\sqrt{2-x}}{\sqrt{x^2+1}}=\frac{0}{\sqrt{5}}=0\)