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I need to prove $$ \lim_{x\rightarrow\ 0}\frac{x^2-8}{{x-8}} =1 $$ using epsilon-delta definition. I know I need to show that for every $\epsilon >0$ there exist a $\delta >0$ such that if $|x|<\delta$, then $| {\frac{x^2-8}{x-8}}-1|<\epsilon$

WHat I did:

$|{\frac{x^2-8}{x-8}}-1|=|{\frac{x^2-x}{x-8}}|$, but Im having troubles now deciding what my delta should be.

I would really appreciate if someone could give me an explanation on how should I choose delta and why. Thanks.

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  • $\begingroup$ What does $x^-8$ mean? It looks like a typo... $\endgroup$ – 5xum Dec 3 '20 at 11:33
  • $\begingroup$ Is $\frac{x-8}{\sqrt{x-8}}$? if so, it is the same as $\sqrt{x-8}$ $\endgroup$ – Tito Eliatron Dec 3 '20 at 11:33
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    $\begingroup$ sorry I had typing error. edited. $\endgroup$ – user114138 Dec 3 '20 at 11:35
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Note that$$\frac{x^2-x}{x-8}=x\frac{x-1}{x-8}.$$Now, if $|x|<1$, then $-1<x<1$ and therefore you have two things:

  • $-9<x-8<-7$, which implies that $|x-8|>7$;
  • $-2<x-1<0$, which implies that $|x-1|<2$.

So,$$|x|<1\implies\left|x\frac{x-1}{x-8}\right|<\frac27|x|.$$Therefore, given $\varepsilon>0$, take $\delta=\min\left\{\frac72\varepsilon,1\right\}$and then\begin{align}|x|<\delta&\implies|x|<\frac72\varepsilon\quad\text{and}\quad|x|<1\\&\implies\left|x\frac{x-1}{x-8}\right|<\varepsilon.\end{align}

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Hint:

$$\left|\frac{x^2-x}{x-8}\right| = |x|\cdot \left|\frac{x-1}{x-8}\right|.$$

Now, if $\delta<1$, then you can further estimate that:

  1. $|x-1| < 3$
  2. $|x-8| > 6$.
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We have

$$\left|x^2-x\over x-8\right|=|x|\left|x-1\over x-8\right|\le|x|{|x|+1\over||x|-8|}$$

Now the main thing that causes a problem is the smallness of the denominator if $|x|\approx8$. But since we're interested in the limit as $x\to0$, it's easy to stay away from $8$ by requiring, say $|x|\le7$, in which case we have

$${|x|+1\over||x|-8|}\le{7+1\over|7-8|}=8$$

So taking $\delta=\min(7,\epsilon/8)$ we have

$$0\lt|x|\lt\delta\implies\left|x^2-x\over x-8\right|\le8|x|\lt8\delta\le\epsilon$$

as desired.

Remark: The key to limit write-ups of this type is to keep in mind that even though we're thinking of small epsilons, we don't want to be embarrassed by claiming an implication that isn't true in case someone decides to use a large $\epsilon$. E.g., if we just let $\delta=\epsilon/8$, we run the risk of the false implication $0\lt|x|\lt8\implies|(x^2-x)/(x-8)|\lt64$, which is falsified, for example, at $x=7.9$ (that is, $0\lt|7.9|\lt8$ but $|(7.9^2-7.9)/(7.9-8)|=545.1\gt64$). The role of the "min" is to avoid such false claims. As the other answers here show, there is no single prefered way to choose your delta; it's somewhat a matter of taste. My preference is to try to isolate what causes things to be uncontrollably large, usually a potential zero in a denominator, impose a restriction brings that term under control, and then look at what happens to everything else.

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We've solved the problem on scrap paper and now present a solution devoid of the process details.

It is easy to show that

$\quad \large |x| \lt \frac{1}{6} \implies |\frac{x-1}{x-8}| \lt \frac{1}{7}$

For $\varepsilon \gt 0$ set $\delta = \text{min}(\frac{1}{6}, 7\varepsilon)$.

If $x \lt \delta$ then

$\quad \large |{\frac{x^2-8}{x-8}-1|=| \frac{x-1}{x-8}}|\,|x| \lt \frac{1}{7} \cdot 7\varepsilon = \varepsilon$

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There are many ways to get an upper bound for $\left| \frac{x-1}{x-8} \right|$ when $x$ is sufficiently small. A couple of things you can do to make this easier: 1) Use the triangle inequality, as one of the answerers did; 2) separate a constant so that you only need to deal with one term involving $x$. So if $|x| < 1$ then

$$\left| \frac{x-1}{x-8} \right|=\left| 1+\frac{7}{x-8} \right| \le 1+\frac{7}{8-x} \le 1+\frac{7}{8-1}=2.$$

Then you have $\delta = \min \left(1, \frac{\varepsilon}{2}\right).$

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