the figure above shows the graph of the twice - differentiable function f and the line tangent to the graph…

Explanation:

Step1: Recall the tangent - line property

Since the line is tangent to (y = f(x)) at ((0,2)), then (f(0)=2) and (f^{\prime}(0)) is the slope of the tangent line.

Step2: Use L'Hopital's Rule

The limit (\lim_{x\rightarrow0}\frac{f(x)e^{-x}-2}{x^{2}-2x}) is in the (\frac{0}{0}) form (because when (x = 0), (f(0)e^{-0}-2=2 - 2=0) and (0^{2}-2\times0 = 0)). By L'Hopital's Rule, (\lim_{x\rightarrow0}\frac{f(x)e^{-x}-2}{x^{2}-2x}=\lim_{x\rightarrow0}\frac{f^{\prime}(x)e^{-x}-f(x)e^{-x}}{2x - 2}).

Step3: Evaluate the new limit

Substitute (x = 0) into (\frac{f^{\prime}(x)e^{-x}-f(x)e^{-x}}{2x - 2}). We know (f(0)=2) and let the slope of the tangent line at (x = 0) be (m=f^{\prime}(0)). Then (\lim_{x\rightarrow0}\frac{f^{\prime}(x)e^{-x}-f(x)e^{-x}}{2x - 2}=\frac{f^{\prime}(0)e^{0}-f(0)e^{0}}{2\times0 - 2}=\frac{f^{\prime}(0)-2}{-2}). From the graph, the slope of the tangent line at ((0,2)) is (m=- 2) (by observing the rise - over - run of the tangent line). Substitute (f^{\prime}(0)=-2) into (\frac{f^{\prime}(0)-2}{-2}), we get (\frac{-2 - 2}{-2}=2).

Answer:

(2)