MHB [ASK] Seemingly Simple Limit Question but I have no Idea

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The limit problem involves calculating $$\lim_{x\to a}\frac{g(x)\cdot f(a)-g(a)\cdot f(x)}{x-a}$$ given specific values for f and g at point a. By applying L'Hôpital's rule due to the 0/0 form when substituting x = a, the derivative of the numerator is found to be g'(x)f(a) - g(a)f'(x). Substituting the known values for g'(a), f(a), g(a), and f'(a) yields a final result of 5. The discussion clarifies that determining the functions f(x) and g(x) is unnecessary for solving the limit. The correct answer to the limit question is 5.
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If f(a) = 2, f'(a) = 1, g(a) = –1, and g'(a) = 2, the value of $$\lim_{x\to a}\frac{g(x)\cdot f(a)-g(a)\cdot f(x)}{x-a}$$ is ...
A. 1
B. 3
C. 5
D. 7
E. 9

$$\lim_{x\to a}\frac{g(x)\cdot f(a)-g(a)\cdot f(x)}{x-a}=\lim_{x\to a}\frac{2g(x)+f(x)}{x-a}$$. How to determine the f(x) and g(x)? And when to use the info that f'(a) = 1 and g'(a) = 2?
 
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You CAN'T "determine the f(x) and g(x)" and you don't need to do. That is not asked.

Taking x= a gives 0/0 so we can use "L'Hopitals rule". The derivative of the numerator is g'(x)f(a)- g(a)f'(x) and the derivative of the denominator is 1 so the limit is $\lim_{x\to a} g'(x)f(a)- g(a)f'(x)= g'(a)f(a)-g(a)f'(a)= (2)(2)- (1)(-1)= 4+ 1= 5$.
 
Ah, so it is indeed a simple question. My bad, Thanks for your help. :)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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