# Using Algebraic Manipulation to evaluate a limit

1. Dec 17, 2011

### EcKoh

1. The problem statement, all variables and given/known data

limx→3 (x2+5x+2) ÷ x2-6x+9

2. The attempt at a solution

You cannot solve through direct substitution because the denominator comes out to equal 0, and you cannot divide by 0. Thus, I factored the lower terms to reach the following:

(x-3)2 or (x-3)(x-3).

However, to my knowledge I cannot do this to the numerator, as ±1 & ±2 do not add up to 5 in any way, shape, or form. As such, I manipulated (x2+5x+2) into the following:

x(x+5)+2

But I am not sure where to go from here as the denominator still adds up to 0, and I am not sure of how to simplify any further. As of right now I have:

x(x+5)+2 ÷ (x-3)(x-3)

2. Dec 17, 2011

### rock.freak667

Try dividing the numerator and denominator by x^2 after dividing the polynomial.

3. Dec 17, 2011

### SammyS

Staff Emeritus
I assume you mean that you're trying to find limx→3 (x2+5x+2) ÷ (x2-6x+9). (The parentheses are important.)

Usually, one simply observes that as x→3, the numerator → 2 and the denominator → 0 (from the positive side) so that the limit is +∞ .

Last edited: Dec 17, 2011