# Calculating limits - general question

• Sep 26th 2010, 08:26 AM
Kakariki
Calculating limits - general question
Hey!
I encountered a question in my homework that was: "Evaluate the limit, if it exists".
$\displaystyle limit as x approaches 2 \frac{x^2 - x + 6}{x-2}$
It is not possible to factor out the denominator of this equation. I tried dividing (x-2) into the equation and got a solution of x+1R8.

My question is: if I can't factor out the denominator of a rational equation, and the limit is asking where that equation would have the denominator at 0 (in the example 2) does this mean the limit does not exist?

I hope I made sense in my question.

Thanks!
• Sep 26th 2010, 08:32 AM
Plato
Yes, it means exactly that. The limit does not exist.

BTW
$$\displaystyle\lim _{x \to 2} \frac{{x^2 - 2x + 6}}{{x - 2}}$$ gives $\displaystyle \displaystyle\lim _{x \to 2} \frac{{x^2 - 2x + 6}}{{x - 2}}$.
• Sep 26th 2010, 10:13 AM
Quote:

Originally Posted by Kakariki
Hey!
I encountered a question in my homework that was: "Evaluate the limit, if it exists".

$\displaystyle \displaystyle\lim_{x \to 2}\frac{x^2 - x + 6}{x-2}$

It is not possible to factor out the denominator of this equation. I tried dividing (x-2) into the equation and got a solution of x+1R8.

My question is: if I can't factor out the denominator of a rational equation, and the limit is asking where that equation would have the denominator at 0 (in the example 2) does this mean the limit does not exist?

I hope I made sense in my question.

Thanks!

You could simplify the fraction a little to

$\displaystyle \displaystyle\frac{x^2-x+6}{x-2}=\frac{x^2-x-2+8}{x-2}=\frac{(x-2)(x+1)+8}{x-2}$

$\displaystyle =\displaystyle\frac{(x-2)(x+1)}{x-2}+\frac{8}{x-2}$

Hence the limit doesn't exist as the second fraction approaches $\displaystyle \pm\infty$ as x approaches 2.
• Sep 27th 2010, 02:58 AM
HallsofIvy
Since you titled this "general question", here is a "general" answer.

To find the limit $\displaystyle \displaystyle\lim_{x\to a}\frac{P(x)}{Q(x)}$, first try just putting x= a. If Q(a) is not 0, the limit is just $\displaystyle \frac{P(a)}{Q(a)}$ (even if P(x)= 0). If Q(a)= 0 but P(a) is NOT 0, then the limit does not exist. If both P(a) and Q(a) are 0, then both P(a) and Q(a) must have a factor of x- a which can be cancelled.