The discussion revolves around the concept of indeterminate forms in calculus, specifically focusing on limits involving logarithmic functions and the application of L'Hopital's Rule. The original poster presents a limit problem involving ln(x-1) and a polynomial expression, questioning why it is not considered an indeterminate form despite yielding a specific value upon direct substitution.
The discussion is ongoing, with participants providing clarifications and examples to illustrate their points. Some participants have offered guidance on when to apply L'Hopital's Rule and the nature of limits that result in undefined expressions. Multiple interpretations of the examples are being explored, indicating a productive exchange of ideas.
Participants are navigating the nuances of limit evaluation, particularly in the context of homework constraints and the need for clear understanding before an upcoming exam. There is an emphasis on the importance of correctly identifying forms and applying appropriate mathematical rules.
The only indeterminate form in your list is 0/0.domyy said: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.
L'Hopital's Rule does not apply in this problem. It can be used on the indeterminate forms [0/0] or [∞/∞].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.
NO - just indeterminate forms [0/0] or [∞/∞].domyy said:So indeterminate form = 0/0
undefined = 2/0
And I apply L'Hopital's rule for both indeterminate and undefined answers?
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:Because I had solved some problems resulting in 6/0 using L'Hopital's rule.
Yes. 0 divided by anything other than 0 is 0.domyy said:0 = 0/2
Is that correct?
domyy said:obs: having these "small" things clarified make ALL the difference. Thanks a lot!
No, because it's not of indeterminate form. It's no 0/0 .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 ?
If you meanIf 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?
How do you figure that? Direct substitution gives 1/0 - 1/0, both of which are undefined.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
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:Should I apply L'Hopital's Rule ?
Is the limit expression (x2 + 2x + 3)/(x - 1)? What you wrote is x2 + 2x + (3/(x - 1)).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.
Yes, that is correct, but you need parentheses around the numerator of your limit expression, like this: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?
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'hospital's rule to a problem and direct substitution results in 1/0, should I write the answer as undefined?