Using Algebraic Manipulation to evaluate a limit

[EcKoh]

Homework Statement

lim_x→3 (x²+5x+2) ÷ x²-6x+92. 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)

 

[rock.freak667]

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

 

[SammyS]

Homework Statement

lim_x→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)

I assume you mean that you're trying to find lim_x→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 +∞ .