0
$\begingroup$

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 = $\frac{22}{1}$
$\endgroup$
6
  • $\begingroup$ @DimitriSurinx I didn't multiply it. I factored both to cancel them out. $\endgroup$ Mar 16, 2013 at 20:52
  • 1
    $\begingroup$ Is the original copied correctly? I get a different $x^2$ term multiplying your denominator. $\endgroup$
    – Mike
    Mar 16, 2013 at 20:53
  • $\begingroup$ @vincentbelkin, my bad sorry. $\endgroup$
    – sxd
    Mar 16, 2013 at 20:53
  • $\begingroup$ @Mike the original question? or in one of my solution? $\endgroup$ Mar 16, 2013 at 20:56
  • 1
    $\begingroup$ Don't be embarrassed, I have done silly things like this more times than I can count. $\endgroup$ Mar 16, 2013 at 21:06

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Unless something was copied wrong, it appears your denominator is equal to

$$(4x^2-13x+4)x-3\ne(4x^2-13x+4)(x-3)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.