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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?

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    @Jordan: Really, if you're interested in a tutor and you don't know of a local resource, I've tutored over Skype in the past. If you're interested, shoot me an email. My email is [very easy to find on my blog](http://mixedmath.wordpress.com/about/) (but not something I post on this forum).2012-06-13

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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.)

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    As for how you'd know to do it, that you just have to know or guess; but in general, if you're interested in the limit $x \to c$ for some constant $c$, the substitution $x = c + \epsilon$ is often useful (because then you can work with the equivalent limit $\epsilon \to 0$, which is often easier).2011-08-27
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$(x^2-6x+5)$ = $(x-1)(x-5)$ 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 $x$:

$5-1 = 4$

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    @Jordan Carlyon: In that case, there would have to be a factor $x$ in the numerator to cancel each other, otherwise the limit might not be defined, in principle, but we would have to see it in detail.2011-08-28
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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:

  1. manipulate algebraically $\frac{P(x)}{Q(x)}=\frac{x^{2}-6x+5}{x-5}$.
  2. 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 $0$.