Ta có: $\lim\limits_{x\to -\sqrt{3}} $`{x^2+3\sqrt{3}}/{3-x^2}` $=\lim\limits_{x\to -\sqrt{3}}$`{x^3+(\sqrt{3})^3}/{(\sqrt{3})^2-x^2}` $=\lim\limits_{x\to -\sqrt{3}}$`{(x+\sqrt{3})(x^2-x\sqrt{3}+3)}/{(\sqrt{3}-x)(\sqrt{3}+x)}` $=\lim\limits_{x\to -\sqrt{3}}$`{x^2-x\sqrt{3}+3}/{\sqrt{3}-x}` `={(-\sqrt{3})^2-(-\sqrt{3}).\sqrt{3}+3}/{\sqrt{3}-(-\sqrt{3})}` `=9/{2\sqrt{3}}=3/2 \sqrt{3}`
\(\lim_{x\to-\sqrt3}\frac{x^3+3\sqrt3}{3-x^2}=\lim_{x\to-\sqrt3}\frac{\left(x+\sqrt3\right)\left(x^2-x\sqrt3+3\right)}{\left(\sqrt3-x\right)\left(\sqrt3+x\right)}\)
\(=\lim_{x\to-\sqrt3}\frac{x^2-x\sqrt3+3}{-x+\sqrt3}=\frac{\left(-\sqrt3\right)^2-\sqrt3\left(-\sqrt3\right)+3}{-\left(-\sqrt3\right)+\sqrt3}\)
\(=\frac{3+3+3}{2\sqrt3}=\frac{9}{2\sqrt3}=\frac{3\sqrt3}{2}\)
