Limit Evaluation: Does $\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}$ Exist?

  • Thread starter fstam2
  • Start date
  • Tags
    Limit
In summary, the limit \lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1} does not exist because there is no common factor between the numerator and denominator. However, we can factor out an (x-1) from the numerator and use synthetic division to find the other factor. This shows that (x-1) must be a common factor since x=1 is a root of both polynomials.
  • #1
fstam2
10
0
Here is the question,
Evaluate this limit:
[tex]\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}[/tex]
since there is no common factor, this limit does not exist, correct?
Maybe I am missing a basic algebra rule for the numerator to find a common factor.
Thank you.
 
Last edited:
Physics news on Phys.org
  • #2
You can factor an (x-1) out of the numerator. You can see that this must be so since x=1 is a root of x^3-1 and a polynomial must factor out into its roots. You can use synthetic division to find what the other factor should be.
 
  • #3
There is no common factor?? If p(1)= 0 for any polynomial p then (x-1) must be a common factor! x3-1= (x-1)(x2+ x+ 1) and x2- 1= (x-1)(x+ 1).
 

FAQ: Limit Evaluation: Does $\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}$ Exist?

1. What is a limit evaluation?

A limit evaluation is a mathematical method used to determine the behavior of a function as the input approaches a specific value. It allows us to determine the value that a function approaches, or "approaches" as the input gets closer and closer to a given value.

2. How do you evaluate a limit?

To evaluate a limit, we plug in values that are close to the given value on either side and observe the behavior of the function. If the function approaches the same value from both sides, then the limit exists and is equal to that value.

3. What is the limit of $\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}$?

The limit of this function as x approaches 1 is equal to 2. This can be evaluated by plugging in values close to 1, such as 1.1 and 0.9, and observing that the function approaches 2 from both sides.

4. Does the limit $\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}$ exist?

Yes, the limit exists and is equal to 2. This can be determined by evaluating the function as x approaches 1 and observing that it approaches the same value from both sides.

5. Why is the limit $\lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}$ equal to 2?

The limit is equal to 2 because as x approaches 1, the function $\frac{x^3-1}{x^2-1}$ simplifies to $\frac{x^2+x+1}{x+1}$, which equals 2 when x is equal to 1. This is also confirmed by evaluating the function from both sides as x approaches 1.

Similar threads

Back
Top