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

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The limit $$\lim_{x\to a}\frac{g(x)\cdot f(a)-g(a)\cdot f(x)}{x-a}$$ evaluates to 5, using L'Hôpital's Rule. Given the values f(a) = 2, f'(a) = 1, g(a) = -1, and g'(a) = 2, the limit simplifies to $$g'(a)f(a)-g(a)f'(a)$$, resulting in (2)(2) - (1)(-1) = 5. This confirms that the problem, initially perceived as complex, is straightforward when applying the correct calculus techniques.

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Monoxdifly
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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'hospital's 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. :)
 

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