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

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

Postby Monox D. I-Fly » Mon Nov 02, 2020 10:38 pm

If f(a) = 2, f'(a) = 1, g(a) = –1, and g'(a) = 2, the value of [tex]\lim_{x\to a}\frac{g(x)\cdot f(a)-g(a)\cdot f(x)}{x-a}[/tex] is ....

A. 1

B. 3

C. 5

D. 7

E. 9


[tex]\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}[/tex]. How to determine the f(x) and g(x)? And when to use the info that f'(a) = 1 and g'(a) = 2?
Monox D. I-Fly
 
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Re: [ASK] Seemingly Simple Limit Question but I have no Idea

Postby Guest » Thu Nov 19, 2020 7:18 pm

Do you not know "L'Hopital's rule"? It says that "If [tex]\lim_{x\to a} F(x)= 0[/tex] and [tex]\lim_{x\to a} G(x)=0[/tex] then [tex]\lim_{x\to a} \frac{F(x)}{G(x)}= \lim_{x\to a} \frac{F'(x)}{G'(a)}[/tex].

Here F(x)= g(x)f(a)- g(a)f(x) and G(x)= x- a so [tex]\lim_{x\to 0} F(x)= F(a)= g(a)f(a)- g(a)f(a)= 0[/tex] and, of course, [tex]g(a)= a- a= 0[/tex].

[tex]F'(x)= g'(x)f(a)- g(a)f'(x)[tex] so [tex]F'(a)= g'(a)f(a)- g(a)f'(a)= (2)(2)-(-1)(1)= 4+ 1= 5[/tex] and, of course, [tex]G'(x)= 1[/tex] so [tex]G'(a)= 1[/tex].
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