# Math Help - Abstraction and Indeterminate Forms of Limits

1. ## Abstraction and Indeterminate Forms of Limits

$\lim_{x\to 0} \frac{2x^2-3x-4}{x^2-1}$

I had this problem on a quiz today and I don't feel comfortable with how I approached it. Since the highest degree of both the variables in the numerator and denominator are the same, I assumed that the limit as x approaches zero would be the quotient of their coefficients, 2. My professor teaches this method of limit taking much more abstractly though when direct substitution doesn't apply, and that's where I am confused. He'll use notation like $\frac{1}{\infty}$ or $-\infty-\infty$ to find the limit of a problem where direct substitution doesn't apply.

For the problem in this post I tried taking a difference of squares in the denominator and factoring the top but that didn't get me anywhere. I guess my question is whether or not anyone's familiar with the abstraction process for indeterminate forms; e.g., $\frac{1}{0}$ being equal to infinity, and if the right answer here is 2?

2. ## Re: Abstraction and Indeterminate Forms of Limits

which limit did you wish to know about?

$\lim_{x \to 0} \frac{2x^2-3x-4}{x^2-1} = 4$

$\lim_{x \to \infty} \frac{2x^2-3x-4}{x^2-1} = 2$

3. ## Re: Abstraction and Indeterminate Forms of Limits

$\lim_{x \to 0}$.

In this case, all that's necessary is direct substitution of zero into the x? I think I'm just looking at limits too hard.

4. ## Re: Abstraction and Indeterminate Forms of Limits

direct substitution works in this particular case.

5. ## Re: Abstraction and Indeterminate Forms of Limits

In a case where it doesn't, specifically for polynomials in rational functions, does that mean it's in an indeterminate form?

yes

7. ## Re: Abstraction and Indeterminate Forms of Limits

Originally Posted by AZach
In a case where it doesn't, specifically for polynomials in rational functions, does that mean it's in an indeterminate form?
Also note that, if P(a)= 0, with P(x) a polynomial, x- a must be a factor. That means that if, setting x= a in both numerator and denominator, both numerator and denominator give 0, you can factor (x- a) out of each and cancel.