a: \(\lim\limits_{x\rightarrow-1}\dfrac{x^3+1}{x+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}\)
\(=\lim\limits_{x\rightarrow-1}x^2-x+1=\left(-1\right)^2-\left(-1\right)+1=2+1=3\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x}-1}{x^3+2x-3}=\dfrac{\sqrt{2}-1}{2^3+2\cdot2-3}=\dfrac{1}{9}\)
c:
\(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{x^3-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{x+1}{x^2+x+1}=\dfrac{1+1}{1+1+1}=\dfrac{2}{3}\)
d:
\(\lim\limits_{x\rightarrow4}\dfrac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim\limits_{x\rightarrow4}\dfrac{\sqrt{x+5}-3}{\sqrt{5-x}-1}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{x+5-9}{\sqrt{x+5}+3}:\dfrac{5-x-1}{\sqrt{5-x}+1}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{x-4}{\sqrt{x+5}+3}\cdot\dfrac{\sqrt{5-x}+1}{-\left(x-4\right)}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{-\sqrt{5-x}-1}{\sqrt{x+5}+3}=\dfrac{-\sqrt{5-4}-1}{\sqrt{4+5}+3}\)
\(=\dfrac{-1-1}{3+3}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
e:
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{2x+3}-1}{\sqrt{x+5}-2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{2x+3-1}{\sqrt{2x+3}+1}:\dfrac{x+5-4}{\sqrt{x+5}+2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{2\left(x+1\right)}{\sqrt{2x+3}+1}\cdot\dfrac{\sqrt{x+5}+2}{x+1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{2\left(\sqrt{x+5}+2\right)}{\sqrt{2x+3}+1}\)
\(=\dfrac{2\left(\sqrt{5-1}+2\right)}{\sqrt{-2+3}+1}=\dfrac{2\left(2+2\right)}{1+1}=4\)
g: \(\lim\limits_{x\rightarrow0^-}\dfrac{1}{x}=-\infty;\lim\limits_{x\rightarrow0^-}2x^2+1=1>0\)
=>\(\lim\limits_{x\rightarrow0^-}\dfrac{2x^2+1}{x}=-\infty\)
