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