# Whats wrong in my answer for this limit $\lim_{x \rightarrow 3} (\frac{2x^3-5x^2-2x-3}{4x^3-13x^2+4x-3})$

The answer is supposed to be 11/17 however I keep getting 22 $$\lim_{x \rightarrow 3} (\frac{2x^3-5x^2-2x-3}{4x^3-13x^2+4x-3})$$

here's my solution:

1. since 3 is a zero of the the denominator it is divisible by x - 3 $$\lim_{x\rightarrow 3} (\frac{2x^3-5x^2-2x-3}{(4x^2-13x+4)(x-3)})$$

2. Used synthetic division by (x -3) on numerator $$\lim_{x\rightarrow 3} (\frac{(2x^2+x+1)(x-3)}{(4x^2-13x+4)(x-3)})$$

3. Cancel (x-3)

What did I misunderstand here? Wolfram alpha isn't able to show the step by step solution so i can't figure it out, however its answer is also 11/17

1. Solve by replacing x by 3 = $\frac{22}{1}$
• @DimitriSurinx I didn't multiply it. I factored both to cancel them out. Mar 16, 2013 at 20:52
• Is the original copied correctly? I get a different $x^2$ term multiplying your denominator.
– Mike
Mar 16, 2013 at 20:53
– sxd
Mar 16, 2013 at 20:53
• @Mike the original question? or in one of my solution? Mar 16, 2013 at 20:56
• Don't be embarrassed, I have done silly things like this more times than I can count. Mar 16, 2013 at 21:06

The denominator indeed is divisible by $x-3$ but the quotient you ended up with isn't quite correct:
$$(4x^2-13x+4)(x-3) = 4 x^3-25 x^2+43 x-12$$
If you used synthetic division you should end up with the factorization $(4 x^2-x+1)(x-3)$, which will give you the correct result.
$$(4x^2-13x+4)x-3\ne(4x^2-13x+4)(x-3)$$