a. \(\mathop {\lim }\limits_{x \to 0} {{\tan 2x} \over {\sin 5x}} = \mathop {\lim }\limits_{x \to 0} {{\sin 2x} \over {\cos 2x.\sin 5x}} \)
\(= \mathop {\lim }\limits_{x \to 0} {{\sin 2x} \over {2x}}.{1 \over {\cos 2x.{{\sin 5x} \over {5x}}}}.{2 \over 5} = {2 \over 5}\)
b. \(\mathop {\lim }\limits_{x \to 0} {{1 - {{\cos }^2}x} \over {x\sin x}} = \mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {2x\sin x\cos x}} = \mathop {\lim }\limits_{x \to 0} {{\sin x} \over {2x\cos x}} = {1 \over 2}\)
c.
\(\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{1 + \sin x - \cos x} \over {1 - \sin x - \cos x}} = \mathop {\lim }\limits_{x \to 0} {{2\sin^2 {x \over 2} + 2\sin {x \over 2}\cos {x \over 2}} \over {2{{\sin }^2}{x \over 2} - 2\sin {x \over 2}\cos {x \over 2}}} \cr & = \mathop {\lim }\limits_{x \to 0} {{\sin {x \over 2} + \cos {x \over 2}} \over {\sin {x \over 2} - \cos {x \over 2}}} = - 1 \cr} \)
