From the graph, it is clear that \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 0\) and \(\mathop {\lim }\limits_{x \to 2} g\left( x \right) = 0\). So, the limit \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}}\) becomes an indeterminate form of type \(\frac{0}{0}\). Now, apply L’Hospital’ Rule to get, \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\) .
From the figure, it can be seen from the graph that \(y = 1.8\left( {x - 2} \right)\) is the tangent line of \(f\) and \(y = \frac{4}{5}\left( {x - 2} \right)\) is the tangent line of \(g\) at point \(\left( {2,0} \right)\). Now, the slope of both the tangent lines at a point on the function is equal to the derivative of the function at the same point. So, it can be obtained as follows:
\(f'\left( 2 \right) = 1.8\)and \(g'\left( 2 \right) = \frac{4}{5}\).
Substitute these values in \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\) to get:
\(\begin{array}{c}\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\\ = \frac{{f'\left( 2 \right)}}{{g'\left( 2 \right)}}\\ = \frac{{1.8}}{{\frac{4}{5}}}\\ = \frac{{\frac{9}{5}}}{{\frac{4}{5}}}\\ = \frac{9}{4}\end{array}\)
Hence, \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{9}{4}\).
