$\begin{array}{l}
a)\mathop {\lim }\limits_{x \to - 3} \dfrac{{1 - {x^2}}}{{x + 1}} = \dfrac{{1 - {{\left( { - 3} \right)}^2}}}{{ - 3 + 1}} = \dfrac{{ - 8}}{{ - 2}} = 4\\
b)\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 2x - 3}}{{x - 1}} = \dfrac{{{3^2} - 2.3 - 3}}{{3 - 1}} = \dfrac{0}{2} = 0\\
c)\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{x + 1}}{{x - 2}} = \dfrac{{3 + 1}}{{3 - 2}} = \dfrac{4}{1} = 4\\
d)\mathop {\lim }\limits_{x \to - {2^ - }} \left( {\sqrt {{x^2} + 5} - 1} \right) = \sqrt {{{\left( { - 2} \right)}^2} + 5} - 1 = 3 - 1 = 2\\
e)\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {3 + \cos 2x} - \sin 4x} \right) = \sqrt {3 + \cos 0} - \sin 0 = 2\\
f)\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x + 1}}{{\cos x - 2}} = \dfrac{{\sin 0 + 1}}{{\cos 0 - 2}} = \dfrac{1}{{1 - 2}} = - 1
\end{array}$
