# Derivative - Limits

• September 5th 2009, 10:33 PM
cokeclassic
Derivative - Limits
Hello, I have a problem with which I seem to be getting nowhere. I have to find the value of the limit, and the answer says that it is 3/2

lim 5x^2-7x+2/x^2-1

lim x-> 1

Thanks for the help.
• September 5th 2009, 10:41 PM
Prove It
Quote:

Originally Posted by cokeclassic
Hello, I have a problem with which I seem to be getting nowhere. I have to find the value of the limit, and the answer says that it is 3/2

lim 5x^2-7x+2/x^2-1

lim x-> 1

Thanks for the help.

Notice that

$5x^2 - 7x + 2 = (5x - 2)(x - 1)$ and $x^2 - 1 = (x + 1)(x - 1)$.

So $\frac{5x^2 - 7x + 2}{x^2 - 1} = \frac{(5x - 2)(x - 1)}{(x + 1)(x - 1)}$

$= \frac{5x - 2}{x + 1}$.

Therefore

$\lim_{x \to 1} \frac{5x^2 - 7x + 2}{x^2 - 1} = \lim_{x \to 1} \frac{5x - 2}{x + 1}$

$= \frac{5(1) - 2}{1 + 1}$

$= \frac{3}{2}$.
• September 5th 2009, 10:42 PM
mr fantastic
Quote:

Originally Posted by cokeclassic
Hello, I have a problem with which I seem to be getting nowhere. I have to find the value of the limit, and the answer says that it is 3/2

lim 5x^2-7x+2/x^2-1

lim x-> 1

Thanks for the help.

Start by noting that $\lim_{x \rightarrow 1} \frac{5x^2 - 7x + 2}{x^2 - 1} = \lim_{x \rightarrow 1} \frac{(5x - 2)(x - 1)}{(x - 1)(x+1)}$
• September 5th 2009, 10:49 PM
Prove It
Quote:

Originally Posted by mr fantastic
Start by noting that $\lim_{x \rightarrow 1} \frac{5x^2 - 7x + 2}{x^2 - 1} = \lim_{x \rightarrow 1} \frac{(5x - 2)(x - 1)}{(x - 1)(x+1)}$

Beat ya :P
• September 6th 2009, 03:48 AM
HallsofIvy
Notice that the reason you could not just set x= 1 and evaluate is because both numerator and denominator are equal to 0 when x= 1. You should then recognize that if a polynomial is 0 at x= a, (x- a) is a factor. Recognizing that x-1 must be a factor of $5x^2- 7x+ 2$ you can just divide it by x-1 to find the other factor.