I'd just like to make sure I'm right about this. I'm pretty sure this limit is unsolvable, or at least not with easy algebra.

$$\lim_{x\to 5}\frac{x^2-5x+6}{x-5}$$

I tried completing the square to factor out the $x-5$ but that renders a fraction. So am I right to say this limit does not exist?

  • 4
    $\begingroup$ You don't even need to factorise it, really, just show that the polynomial in the numerator evaluates to something other than $0$ at $x=5$. $\endgroup$ – G. H. Faust Jun 16 '14 at 4:06
  • $\begingroup$ Why would you complete the square? The numerator immediately factors as $(x-2)(x-3)$. Use rational roots test if it isn't immediate for you... $\endgroup$ – Eric Towers Jun 16 '14 at 4:08
  • $\begingroup$ Are you asked to prove that the limit doesn't exist? $\endgroup$ – 3x89g2 Jun 16 '14 at 4:11
  • $\begingroup$ Misakov: No, just note if it does not. $\endgroup$ – Irresponsible Newb Jun 16 '14 at 4:13
  • $\begingroup$ G. H. Faust: Oh yeah, that should have been obvious to me. $\endgroup$ – Irresponsible Newb Jun 16 '14 at 4:14

$$\lim_{x\to 5^+}\frac{x^2-5x+6}{x-5}=\frac{6}{0^+}=+\infty$$ $$\lim_{x\to 5^-}\frac{x^2-5x+6}{x-5}=\frac{6}{0^-}=-\infty$$

So limit does not exist!



If $x-5\ne0,$ $$\frac{x^2-5x+6}{x-5}=x+\frac6{x-5}$$

Now if $x\to5,x\ne5$


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.