# Can I solve this limit?

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?

• You don't even need to factorise it, really, just show that the polynomial in the numerator evaluates to something other than $0$ at $x=5$. – G. H. Faust Jun 16 '14 at 4:06
• 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... – Eric Towers Jun 16 '14 at 4:08
• Are you asked to prove that the limit doesn't exist? – 3x89g2 Jun 16 '14 at 4:11
• Misakov: No, just note if it does not. – Irresponsible Newb Jun 16 '14 at 4:13
• G. H. Faust: Oh yeah, that should have been obvious to me. – Irresponsible Newb Jun 16 '14 at 4:14

$$\lim_{x\to 5^+}\frac{x^2-5x+6}{x-5}=\frac{6}{0^+}=+\infty$$ $$\lim_{x\to 5^-}\frac{x^2-5x+6}{x-5}=\frac{6}{0^-}=-\infty$$
If $x-5\ne0,$ $$\frac{x^2-5x+6}{x-5}=x+\frac6{x-5}$$
Now if $x\to5,x\ne5$