3

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?

Américo Tavares
  • 39,180

3 Answers3

4

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 $0$.

Américo Tavares
  • 39,180
3

$(x^2-6x+5)$ = $(x-1)(x-5)$ 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 $x$:

$5-1 = 4$

Dair
  • 3,278
  • How do I know to factor it to that instead of something else? –  Aug 27 '11 at 22:46
  • It's called factoring a quadratic, and the result is unique, up to constants. – The Chaz 2.0 Aug 27 '11 at 22:47
  • Oh is that some stuff I need to memorize? –  Aug 27 '11 at 22:49
  • @Jordan: It is called factoring. In this case quadratic factoring: http://www.purplemath.com/modules/factquad.htm – Dair Aug 27 '11 at 22:54
  • Oh okay, I was just wondering if that was the quadratic formula that was used. –  Aug 27 '11 at 22:55
  • @Jordan You can "factor without factoring". If you use the quadratic equation, you would find that the roots of $x^2 - 6x + 5$ are $x = 1$ and $x = 5$. This tells you that $x^2 - 6x + 5 = (x - 1)(x - 5)$ without having to do factoring as you usually think of it. – Austin Mohr Aug 27 '11 at 22:57
  • You've written division where you wanted multiplication, anon, you might want to edit. – Gerry Myerson Aug 27 '11 at 23:00
  • @Gerry: Yes thank you, my bad. – Dair Aug 27 '11 at 23:01
  • @Jordan, if you're happier with division than with factorization you might take the suggestion Bill made in the comments on the question. – Gerry Myerson Aug 27 '11 at 23:01
  • I am familiar with this technique, it has just been a long time since I have factored. –  Aug 27 '11 at 23:09
  • "you don't have to worry about dividing by zero" because it's a limiting process, and actually one never divides by zero, when $x$ tends to $5$ ($x−5$ approaches zero but is never equal to 0). – Américo Tavares Aug 27 '11 at 23:43
  • But what if x approaches 0 and the denominator is x? –  Aug 27 '11 at 23:49
  • @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. – Américo Tavares Aug 28 '11 at 01:11
1

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.)

Ilmari Karonen
  • 26,841
  • Why can x = 5+ e? –  Aug 27 '11 at 23:23
  • @Jordan Carlyon: Why can't it? More seriously, for any real number $x$, we can write it in the form $x = 5 + \epsilon$, where $\epsilon = x - 5$. It's also then easy to show that $\epsilon = 0$ if and only if $x = 5$. – Ilmari Karonen Aug 27 '11 at 23:27
  • What is e? Or is that just a variable like x or y? –  Aug 27 '11 at 23:29
  • @Jordan Carlyon: Yes, it's just a variable. (Mathematicians often like to use the Greek letter epsilon ($\epsilon$) for "very small" quantities — especially ones which tend to zero in some limit we're interested in. But that's just a matter of tradition; any other symbol would work just as well.) – Ilmari Karonen Aug 27 '11 at 23:34
  • But I don't understand why or how that is being used and what the purpose of it is. I have never seen anything like that before, why would I want to do that? Why can I just change a variable to a limit plus a different variable? What allows me to do that and how do I know that it if beneficial to solving the problem without guessing? –  Aug 27 '11 at 23:38
  • @Jordan Carlyon: The reason you can do it is the substitution property of equality: if $a=b$, then you can take any expression containing $a$ (as a free variable) and replace it with $b$ without changing the value of the expression. This holds even if $a$ and/or $b$ are arbitrarily complex expressions instead of just single variables. – Ilmari Karonen Aug 27 '11 at 23:47
  • 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). – Ilmari Karonen Aug 27 '11 at 23:52