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.

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- Sep 5th 2009, 10:33 PMcokeclassicDerivative - 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. - Sep 5th 2009, 10:41 PMProve It
Notice that

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

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

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

Therefore

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

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

$\displaystyle = \frac{3}{2}$. - Sep 5th 2009, 10:42 PMmr fantastic
- Sep 5th 2009, 10:49 PMProve It
- Sep 6th 2009, 03:48 AMHallsofIvy
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 $\displaystyle 5x^2- 7x+ 2$ you can just divide it by x-1 to find the other factor.