Finding the limit of a quotient I am trying to find the limit of $(x^2-6x+5)/(x-5)$ as it approaches $5$.
I assume that I just plug in $5$ for $x$ and for that I get $0/0$ but my book says $4$. I try and factor and I end up with $(25-30+5)/(5-5)$ which doesnt seem quite right to me but I know that if I factor out $5$ and get rid of the $5-5$ (although that would make it $1-1$ wouldn't it?) that leaves me with $5-6+5$ which is $4$.
What do I need to do in this problem?
 A: Let $P(x)=x^{2}-6x+5$ and $Q(x)=x-5$. Since $P(x)$ and $Q(x)$ are continuous
and $P(5)=Q(5)=0$, $\frac{P(5)}{Q(5)}$ is undetermined. You have two
alternatives: 


*

*manipulate algebraically $\frac{P(x)}{Q(x)}=\frac{x^{2}-6x+5}{x-5}$. 

*use L'Hôpital's rule
$$\lim_{x\rightarrow 5}\frac{P(x)}{Q(x)}=\lim_{x\rightarrow 5}\frac{P^{\prime
}(x)}{Q^{\prime }(x)}=\lim_{x\rightarrow 5}\frac{2x-6}{1}=2\cdot 5-6=4.$$


In option 1, since $P(5)=0$, you know that you can factor $P(x)$ as $$P(x)=x^{2}-6x+5=(x-5)(x-c).$$
You can compute $c=1$, by solving the equation 
$$x^{2}-6x+5=0.$$ 
Instead you can perform a long division, as suggested by Bill Dubuque, to evaluate $P(x)/Q(x)=x-1$.
So, $$P(x)=x^{2}-6x+5=(x-5)(x-1)$$ and $$\lim_{x\rightarrow 5}%
\frac{P(x)}{Q(x)}=\lim_{x\rightarrow 5}\frac{(x-5)(x-1)}{x-5}%
=\lim_{x\rightarrow 5}(x-1)=5-1=4.$$ 
You are allowed to divide $P(x)$ and $Q(x)$ by $x-5$, because you perform a limiting process, and you actually never make $x=5$, which means $x-5$ is never equal to $0$.
A: $(x^2-6x+5)$ = $(x-1)(x-5)$ cancel out the $x-5$ and you don't have to worry about dividing by zero.
$x-1$ is the end result. Plug in $5$ for $x$:
$5-1 = 4$
A: Anyway, you can also do it from first principles without any factoring tricks:
Let $x = 5 + \epsilon$.  Then, when $x \ne 5$, and thus $\epsilon \ne 0$,
$$\begin{aligned}
\frac{x^2 - 6x + 5}{x-5}
&= \frac{(5 + \epsilon)^2 - 6(5 + \epsilon) + 5}{5 + \epsilon - 5} \\
&= \frac{(25 + 10\epsilon + \epsilon^2) - (30 + 6\epsilon) + 5}{\epsilon} \\
&= \frac{4\epsilon + \epsilon^2}{\epsilon} = 4 + \epsilon.
\end{aligned}$$
We thus see that
$$\lim_{x \to 5} \frac{x^2 - 6x + 5}{x-5} = \lim_{\epsilon \to 0}\, 4 + \epsilon = 4.$$
(Addendum: The reason for choosing that particular substitution is simple: we want to know what happens when $x$ gets close to $5$; $\epsilon = x - 5$ tells how close $x$ is to $5$.  In particular, if the limit as $x \to 5$ is well defined, then after the substitution and simplification we should end up with the limit plus some terms that vanish as $\epsilon \to 0$, as we indeed do.  If, instead, we ended up with some terms like $1/\epsilon$ that diverge as $\epsilon \to 0$, then we'd know that the limit was not well defined.)
