# Indeterminate Forms and L'Hopital's Rule

• domyy
As you can see, it depends on the situation and how you get to the 1/0. In general, if you end up with 1/0 after applying L'Hopital's Rule, you should check the limit from both sides to see if it approaches ∞ or -∞, or if it is undefined.
domyy

## Homework Statement

lim ln(x-1)/(x2-x-4)
x->2

## The Attempt at a Solution

Well, I thought that every time I had answers as 0/0, 2/0 or 0/2, for instance, they would constitute as indeterminate forms.

I have the answer sheet for this problem. It says "answer: 0/-2 = 0".
Why isn't it considerate indeterminate form?

My prof. didnt use L'Hopital's Rule here but direct substitution.

domyy said:

## Homework Statement

lim ln(x-1)/(x2-x-4)
x->2

## The Attempt at a Solution

Well, I thought that every time I had answers as 0/0, 2/0 or 0/2, for instance, they would constitute as indeterminate forms.
The only indeterminate form in your list is 0/0.

2/0 is undefined (which is different), and 0/2 = 0.
domyy said:
I have the answer sheet for this problem. It says "answer: 0/-2 = 0".
Why isn't it considerate indeterminate form?

My prof. didnt use L'Hopital's Rule here but direct substitution.
L'Hopital's Rule does not apply in this problem. It can be used on the indeterminate forms [0/0] or [∞/∞].

So indeterminate form = 0/0
undefined = 2/0

And I apply L'Hopital's rule for both indeterminate and undefined answers?

Because I had solved some problems resulting in 6/0 using L'Hopital's rule.

0 = 0/2

Is that correct?

obs: having these "small" things clarified make ALL the difference. Thanks a lot! This forum is crucial in my studies. I have an exam approaching and need to have all these info clear in my mind.

Last edited:
domyy said:
So indeterminate form = 0/0
undefined = 2/0

And I apply L'Hopital's rule for both indeterminate and undefined answers?
NO - just indeterminate forms [0/0] or [∞/∞].
domyy said:
Because I had solved some problems resulting in 6/0 using L'Hopital's rule.
L'Hopital's Rule can result in a value that is undefined, but you can't use it on an expression that is undefined. Do you understand the difference in what I wrote?
domyy said:
0 = 0/2

Is that correct?
Yes. 0 divided by anything other than 0 is 0.
domyy said:
obs: having these "small" things clarified make ALL the difference. Thanks a lot!

I'd like to get three examples to make sure I understood, if you don't mind. Here it is:

EXAMPLE 1
Lim [1/x - 1/(x2)]
x-> 0

Direct substitution will give me -1/0

Should I apply L'Hopital's Rule ?

If so, I'll have 0/2.
The book says the answer is ∞

EXAMPLE 2

Lim x2 + 2x + 3/(x-1)
x->∞

I applied L'Hopital's Rule here twice and got 2/0.

My book says the answer should be ∞

EXAMPLE 3

Lim 3x2 - 2x + 1/(2x2 + 3)
x->∞

Applying L'Hopital's rule twice will give me 3/2. And that's the same answer the book gives me. So I guess that's right?

domyy said:
I'd like to get three examples to make sure I understood, if you don't mind. Here it is:

EXAMPLE 1
Lim [1/x - 1/(x2)]
x-> 0

Direct substitution will give me -1/0

Should I apply L'Hopital's Rule ?
No, because it's not of indeterminate form. It's no 0/0 .

B.T.W: The result is -∞ not ∞ .
If so, I'll have 0/2.
The book says the answer is ∞

EXAMPLE 2

Lim x2 + 2x + 3/(x-1)
x->∞
If you mean
Lim (x2 + 2x + 3)/(x-1) ,
x->∞

then it is indeterminate, of the form ∞/∞ .

After applying L'Hôpital's rule once it will be of the form ∞/1, thus not indeterminate.
I applied L'Hopital's Rule here twice and got 2/0.

My book says the answer should be ∞

EXAMPLE 3

Lim (3x2 - 2x + 1)/(2x2 + 3)
x->∞

Applying L'Hopital's rule twice will give me 3/2. And that's the same answer the book gives me. So I guess that's right?

domyy said:
I'd like to get three examples to make sure I understood, if you don't mind. Here it is:

EXAMPLE 1
Lim [1/x - 1/(x2)]
x-> 0

Direct substitution will give me -1/0
How do you figure that? Direct substitution gives 1/0 - 1/0, both of which are undefined.
domyy said:
Should I apply L'Hopital's Rule ?
No. L'Hopital's Rule applies only to limits of the form f(x)/g(x). It doesn't apply to sums, differences, or products.
domyy said:
If so, I'll have 0/2.
The book says the answer is ∞

EXAMPLE 2

Lim x2 + 2x + 3/(x-1)
x->∞

I applied L'Hopital's Rule here twice and got 2/0.
Is the limit expression (x2 + 2x + 3)/(x - 1)? What you wrote is x2 + 2x + (3/(x - 1)).

Apply L'H once to get your limit.
domyy said:
My book says the answer should be ∞

EXAMPLE 3

Lim 3x2 - 2x + 1/(2x2 + 3)
x->∞

Applying L'Hopital's rule twice will give me 3/2. And that's the same answer the book gives me. So I guess that's right?
Yes, that is correct, but you need parentheses around the numerator of your limit expression, like this:
Lim (3x2 - 2x + 1)/(2x2 + 3)
x->∞

=D Thank you both so much for your clear explanations. That will prevent me from getting wrong answers from now on. I'll pay attention to these rules when applying L'Hopital's

Just a question to sum it all up:

Based on what I learned here, L'hopital's rule does not apply to sum, differences and subtractions. It also only applies to answers in the indeterminate form of 0/0 and infinity/infinity. However, you said that 1/0 is undefined. So if I am supposed to apply l'hopitals rule to a problem and direct substitution results in 1/0, should I write the answer as undefined?

domyy said:
Just a question to sum it all up:

Based on what I learned here, L'hopital's rule does not apply to sum, differences and subtractions. It also only applies to answers in the indeterminate form of 0/0 and infinity/infinity. However, you said that 1/0 is undefined. So if I am supposed to apply l'hopitals rule to a problem and direct substitution results in 1/0, should I write the answer as undefined?

If you apply L'Hopital's Rule and end up with something of the form 1/0, the answer could be ∞, -∞, or undefined.

Some examples:
$$\lim_{x \to 0} \frac{1}{x^2} = ∞$$
$$\lim_{x \to 0^+} \frac{1}{x} = ∞$$
$$\lim_{x \to 0^-} \frac{1}{x} = -∞$$
$$\lim_{x \to 0} \frac{1}{x} ~~ \text{is undefined}$$

## What are indeterminate forms?

Indeterminate forms are mathematical expressions that cannot be evaluated in a straightforward manner. They arise when attempting to find the limit of a function or expression, and the result is undefined or takes on an infinite value.

## What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical theorem that provides a method for evaluating limits of indeterminate forms. It states that the limit of a quotient of two functions is equal to the limit of their derivatives, under certain conditions.

## When should I use L'Hopital's Rule?

L'Hopital's Rule should only be used when attempting to evaluate the limit of an indeterminate form. This means that the numerator and denominator of the expression both approach 0, infinity, or negative infinity, and the limit is undefined.

## What are the conditions for using L'Hopital's Rule?

To use L'Hopital's Rule, the limit must be of the form 0/0 or infinity/infinity. Additionally, the functions in the numerator and denominator must be differentiable in a neighborhood around the limit point.

## Can L'Hopital's Rule be used for all indeterminate forms?

No, L'Hopital's Rule can only be used for the indeterminate forms of 0/0 and infinity/infinity. Other indeterminate forms, such as 0*infinity or infinity - infinity, require different methods for evaluation.

• Calculus and Beyond Homework Help
Replies
1
Views
719
• Calculus and Beyond Homework Help
Replies
14
Views
1K
• Calculus and Beyond Homework Help
Replies
23
Views
1K
• Calculus and Beyond Homework Help
Replies
34
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
24
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
35
Views
3K