Understanding L'Hôpital's rule

In summary, if you are trying to find the limit of a derivative, you need to use the boundary rules and the definition of the derivative, and justify the limit using the information you have.
  • #1
Kolika28
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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!
 
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  • #2
Kolika28 said:
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!
 
  • #3
Mark44 said:
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)} ##
 
  • #4
Kolika28 said:
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.
 
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  • #5
Mark44 said:
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?
 
  • #6
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)}##?
Kolika28 said:
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.
 
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  • #7
I
Mark44 said:
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!
 

What is L'Hôpital's rule?

L'Hôpital's rule is a mathematical tool used to evaluate the limit of a function involving indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of the quotient of two functions is indeterminate, then the limit of the quotient of their derivatives will be the same.

When should L'Hôpital's rule be used?

L'Hôpital's rule should only be used when the limit of a function is indeterminate. It should not be used for all limits, as it may give incorrect results.

How do I apply L'Hôpital's rule?

To apply L'Hôpital's rule, you must first take the derivative of the numerator and denominator of the function. Then, you can evaluate the limit of the quotient of the derivatives. If the limit is still indeterminate, you can repeat the process until a non-indeterminate value is obtained.

What are the common mistakes when using L'Hôpital's rule?

One common mistake when using L'Hôpital's rule is not checking if the limit is truly indeterminate. Sometimes, the limit can be evaluated using other methods. Another mistake is taking the derivative incorrectly, which can lead to incorrect results.

Are there any limitations to L'Hôpital's rule?

Yes, L'Hôpital's rule only applies to limits involving indeterminate forms. It cannot be used for limits that are not indeterminate. Additionally, it may not work for complex functions or functions with multiple variables.

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