# Help with L'Hôpital's rule

## Homework Statement:

Let f and g be derivable functions and let a be a real number such that

##f(a)=g(a)=0 ##
##g'(a) ≠ 0 ##

Justify that
##\frac{f'(a)}{g'(a)} ## = ##\lim_{x\to a}\frac{f(x)}{g(x)}##

You may only use the definition of the derivative and boundary rules.

## Relevant Equations:

##\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}##
My attempt:
##\frac{f'(a)}{g'(a)} ## =
##\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\cdot\frac{h}{g(a+h)-g(a)}##
= ##\lim_{h\to 0}\frac{f(a+h)-f(a)}{g(a+h)-g(a)}##

I don't think I am doing this right. I don't even understand how I am supposed to use the boundary rules. I really appreciate some help!

Delta2

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Mark44
Mentor
Homework Statement: Let f and g be derivable functions and let a be a real number such that
In English we say "differentiable" not "derivable."
Kolika28 said:
##f(a)=g(a)=0 ##
##g'(a) ≠ 0 ##

Justify that
##\frac{f'(a)}{g'(a)} ## = ##\lim_{x\to a}\frac{f(x)}{g(x)}##

You may only use the definition of the derivative and boundary rules.
Homework Equations: ##\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}##

My attempt:
##\frac{f'(a)}{g'(a)} ## =
##\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\cdot\frac{h}{g(a+h)-g(a)}##
No, this (above) isn't correct.
##\frac{f'(a)}{g'(a)} ##
##= \frac{\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}}{\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}}##
You would need to justify turning this quotient of limits into the limit of a quotient (or the limit of the product you wrote.)

As a hint, you know that the limit in the denominator exists, right? You also know the values of f(a) and g(a).
Kolika28 said:
= ##\lim_{h\to 0}\frac{f(a+h)-f(a)}{g(a+h)-g(a)}##

I don't think I am doing this right. I don't even understand how I am supposed to use the boundary rules. I really appreciate some help!

In English we say "differentiable" not "derivable."
No, this (above) isn't correct.
##\frac{f'(a)}{g'(a)} ##
##= \frac{\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}}{\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}}##
You would need to justify turning this quotient of limits into the limit of a quotient (or the limit of the product you wrote.)

As a hint, you know that the limit in the denominator exists, right? You also know the values of f(a) and g(a).
Hmm, I'm sorry, but I takes some time for me to understand. So I can't write the limit as a product of those two fractions to get rid of the h? I know that the limit in the denominator exist and the values of f(a) and g(a) is equal to zero, but I don't see how I am supposed to use those facts to get to the term

##\lim_{x\to a}\frac{f(x)}{g(x)} ##

Mark44
Mentor
Hmm, I'm sorry, but I takes some time for me to understand. So I can't write the limit as a product of those two fractions to get rid of the h?
Under what conditions is it true that ##\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}##? You seemed to gloss over this by assuming that it's always true.
Kolika28 said:
I know that the limit in the denominator exist and the values of f(a) and g(a) is equal to zero, but I don't see how I am supposed to use those facts to get to the term

##\lim_{x\to a}\frac{f(x)}{g(x)} ##
##\lim_{x \to a} F(x)## is the same as ##\lim_{h \to 0} F(a + h)##, right? You can change to a different variable for the limit.

Delta2
Under what conditions is it true that limx→af(x)g(x)=limx→af(x)limx→ag(x)limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}? You seemed to gloss over this by assuming that it's always true.
If the function is continuous and differentiable?

Mark44
Mentor
Mark44 said:
Under what conditions is it true that ##\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}##?
If the function is continuous and differentiable?
Differentiability isn't necessary. If both limits on the right side of the equation I wrote exist, and the limit in the denominator is not zero, then the quotient of the limits equals the limit of the quotients.

hutchphd and Delta2
I
Differentiability isn't necessary. If both limits on the right side of the equation I wrote exist, and the limit in the denominator is not zero, then the quotient of the limits equals the limit of the quotients.
I have been looking at the information you have given me, and I finally understand it! Thank you so much!