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

[Monoxdifly]

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?

 

[HOI]

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 $\lim_{x\to a} g'(x)f(a)- g(a)f'(x)= g'(a)f(a)-g(a)f'(a)= (2)(2)- (1)(-1)= 4+ 1= 5$.

 

[Monoxdifly]

Ah, so it is indeed a simple question. My bad, Thanks for your help. :)