Prove limit x approaches 0 of a rational function = ratio of derivatives

[FlorenceC]

1. The problem statement, all variables and given/known dat
If f and g are differentiable functions with f(O) = g(0) = 0 and g'(O) not equal 0, show that
lim f(x) = f'(0)
x->0 g(x) g'(0)

The Attempt at a Solution

I know that lim as x→a f(a) = f(a) if function is continuous. since its differentiable it's continuous. so lim x→0 f(x) = f(0). and lim x→0 g(x) = g(0) but you can't have a 0/0.
I have so idea how to get to the derivative part.

 

[PeroK]

1. The problem statement, all variables and given/known dat
If f and g are differentiable functions with f(O) = g(0) = 0 and g'(O) not equal 0, show that
lim f(x) = f'(0)
x->0 g(x) g'(0)

The Attempt at a Solution

I know that lim as x→a f(a) = f(a) if function is continuous. since its differentiable it's continuous. so lim x→0 f(x) = f(0). and lim x→0 g(x) = g(0) but you can't have a 0/0.
I have so idea how to get to the derivative part.

This is essentially a special case of L'Hospital's rule. Do you know the Mean Value Theorem?

 

[FlorenceC]

This is essentially a special case of L'Hospital's rule. Do you know the Mean Value Theorem?

Yes. but how does mvt apply? we didn't learn lhopitals rule yet.

 

[PeroK]

Yes. but how does mvt apply? we didn't learn lhopitals rule yet.

Actually, you don't even need the MVT. Just think about the definition of a derivative.

 

[RUber]

This is one of those cases where adding zero back into the expression makes it more clear:
$\lim_{a\to 0} f(a) = \lim_{a\to 0} f(0+a)-f(0)$ ...