Answer
$$\left. {\text{a}} \right)0.1,{\text{ }}\left. {\text{b}} \right)\frac{1}{{10}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{\sin \left( {x - 5} \right)}}{{{x^2} - 25}} \cr
& \left. {\text{a}} \right){\text{Completing the table}} \cr
& f\left( 4 \right) = \frac{{\sin \left( {4 - 5} \right)}}{{{4^2} - 25}} = 0.0934 \cr
& f\left( {4.5} \right) = \frac{{\sin \left( {4.5 - 5} \right)}}{{{{4.5}^2} - 25}} = 0.1009 \cr
& f\left( {4.9} \right) = \frac{{\sin \left( {4.9 - 5} \right)}}{{{{4.9}^2} - 25}} = 0.1008 \cr
& f\left( {5.1} \right) = \frac{{\sin \left( {5.1 - 5} \right)}}{{{{5.1}^2} - 25}} = 0.0988 \cr
& f\left( {5.5} \right) = \frac{{\sin \left( {5.5 - 5} \right)}}{{{{5.5}^2} - 25}} = 0.0913 \cr
& f\left( 6 \right) = \frac{{\sin \left( {6 - 5} \right)}}{{{6^2} - 25}} = 0.0764 \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
x&4&{4.5}&{4.9}&{5.1}&{5.5}&6 \\
{f\left( x \right)}&{0.0934}&{0.1009}&{0.1008}&{0.0988}&{0.0913}&{0.0764}
\end{array}}\]
$$\eqalign{
& {\text{Then we can assume that}} \cr
& \mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {x - 5} \right)}}{{{x^2} - 25}} = 0.1 \cr
& \cr
& \left. {\text{b}} \right){\text{Calculating the exact value of the limit}} \cr
& \mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {x - 5} \right)}}{{{x^2} - 25}} = \mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} \cr
& {\text{Using the property of product of limits}} \cr
& \mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {x - 5} \right)}}{{{x^2} - 25}} = \mathop {\lim }\limits_{x \to 5} \frac{1}{{\left( {x + 5} \right)}}\mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {x - 5} \right)}}{{\left( {x - 5} \right)}} \cr
& {\text{Calculate}} \cr
& \mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {x - 5} \right)}}{{{x^2} - 25}} = \left( {\frac{1}{{5 + 5}}} \right)\left( 1 \right) \cr
& \mathop {\lim }\limits_{x \to 5} \frac{{\sin \left( {x - 5} \right)}}{{{x^2} - 25}} = \frac{1}{{10}} \cr} $$
