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

Click For Summary
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.
Monoxdifly
MHB
Messages
288
Reaction score
0
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?
 
Mathematics news on Phys.org
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. :)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K