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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?

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

So limit does not exist!

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HINT:

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

Now if $x\to5,x\ne5$

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