Answer
$$\eqalign{
& \left( {\bf{a}} \right)1 \cr
& \left( {\bf{b}} \right)\frac{2}{3} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\bf{a}} \right)\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{\tan x}} \cr
& {\text{Evaluate the given limit without using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{\tan x}} = \frac{{\sin 0}}{{\tan 0}} = \frac{0}{0} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{\tan x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{\frac{{\sin x}}{{\cos x}}}} = \mathop {\lim }\limits_{x \to 0} \cos x \cr
& \mathop {\lim }\limits_{x \to 0} \cos x = \cos 0 = 1 \cr
& {\text{Evaluate using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{\tan x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {\sin x} \right]}}{{\frac{d}{{dx}}\left[ {\tan x} \right]}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{{{\sec }^2}x}} = \mathop {\lim }\limits_{x \to 0} {\cos ^3}x \cr
& \mathop {\lim }\limits_{x \to 0} {\cos ^3}x = {\cos ^3}\left( 0 \right) = 1 \cr
& \cr
& \left( {\bf{b}} \right)\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^3} - 1}} \cr
& {\text{Evaluate the given limit without using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^3} - 1}} = \frac{0}{0} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^3} - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{x + 1}}{{{x^2} + x + 1}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{x + 1}}{{{x^2} + x + 1}} = \frac{{1 + 1}}{{{1^2} + 1 + 1}} = \frac{2}{3} \cr
& {\text{Evaluate using L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^3} - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left[ {{x^2} - 1} \right]}}{{\frac{d}{{dx}}\left[ {{x^3} - 1} \right]}} = \mathop {\lim }\limits_{x \to 1} \frac{{2x}}{{3{x^2}}} = \mathop {\lim }\limits_{x \to 1} \frac{2}{{3x}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{2}{{3x}} = \frac{2}{{3\left( 1 \right)}} = \frac{2}{3} \cr} $$
