Indeterminate Forms and L'Hopital's Rule

[domyy]

Homework Statement

lim ln(x-1)/(x²-x-4)
x->2

Homework Equations

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.

 

[Mark44]

Homework Statement

lim ln(x-1)/(x²-x-4)
x->2

Homework Equations

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.

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 [∞/∞].

 

[domyy]

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.

 

[Mark44]

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 [∞/∞].

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?

0 = 0/2

Is that correct?

Yes. 0 divided by anything other than 0 is 0.

obs: having these "small" things clarified make ALL the difference. Thanks a lot!

 

[domyy]

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?

 

[SammyS]

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?

 

[Mark44]

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.

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.

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.

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->∞

 

[domyy]

=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

 

[domyy]

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'hospital's rule to a problem and direct substitution results in 1/0, should I write the answer as undefined?

 

[Mark44]

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