Understanding L'Hôpital's rule

[Kolika28]

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!

 

[Mark44]

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."

$f(a)=g(a)=0$
$g'(a) ≠ 0$

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).

= $\lim_{h\to 0}\frac{f(a+h)-f(a)}{g(a+h)-g(a)}$

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!

 

[Kolika28]

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]

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.

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.

 

[Kolika28]

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]

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.

 

[Kolika28]

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!