\(a,\lim\limits_{x\rightarrow1}\dfrac{3x-1}{\left(x-1\right)^2}=\)\(\lim\limits_{x\rightarrow1}\dfrac{3x-1}{x^2-2x+1}=\)\(\lim\limits_{x\rightarrow1}\dfrac{x\left(3-\dfrac{1}{x}\right)}{x^2\left(1-\dfrac{2}{x}+\dfrac{1}{x^2}\right)}=\)\(\lim\limits_{x\rightarrow1}\dfrac{\left(3-\dfrac{1}{x}\right)}{x\left(1-\dfrac{2}{x}+\dfrac{1}{x^2}\right)}=+\infty\)
\(b,\lim\limits_{x\rightarrow-3}\dfrac{2x^2+5x-3}{\left(x+3\right)^2}=\)\(\lim\limits_{x\rightarrow-3}\dfrac{2x^2+5x-3}{x^2+6x+9}=\)\(\lim\limits_{x\rightarrow-3}\dfrac{x^2\left(2+\dfrac{5}{x}-\dfrac{3}{x^2}\right)}{x^2\left(1+\dfrac{6}{x}+\dfrac{9}{x^2}\right)}=\)\(\lim\limits_{x\rightarrow-3}\dfrac{2+\dfrac{5}{x}-\dfrac{3}{x^2}}{1+\dfrac{6}{x}+\dfrac{9}{x^2}}=+\infty\)
\(\lim\limits_{x\rightarrow1}\dfrac{3x-1}{\left(x-1\right)^2}=\dfrac{3.1-1}{0}=+\infty\)
b.
\(\lim\limits_{x\rightarrow-3}\dfrac{2x^2+5x-3}{\left(x+3\right)^3}=\lim\limits_{x\rightarrow-3}\dfrac{\left(2x-1\right)\left(x+3\right)}{\left(x+3\right)^3}\)
\(=\lim\limits_{x\rightarrow-3}=\dfrac{2x-1}{\left(x+3\right)^2}=\dfrac{-7}{0}=-\infty\)
