# Using $\epsilon-\delta$ proof to prove continuity

Use an $\epsilon-\delta$ proof to show that $f : R \setminus \left \{ \frac{-3}{2} \right \} \rightarrow R$ , $$f(x) = \frac{3x^2-2x-5}{2x+3}$$ is continuous at $x = -1$

Hello there. Can anyone here help me with this? I know i need to show that $|x-l|<\delta$ implies $|f(x) - f(l)| < \epsilon$ but I don't know how to do it for this specific example.

• Is there any reason in particular that this example is difficult/interesting for you? – Andres Mejia May 14 '16 at 15:14
• Start with $|f(x)-f(l)|<\epsilon$. Then, try to have a sequence of inequalities so that you end up with something like $|x-l|<$ "something with $\epsilon$". Then, choose $\delta$ equals that "something with $\epsilon$" – H. Potter May 14 '16 at 15:16

Let $\epsilon>0$. We want to pick a $\delta$ such that $|x-(-1)|=|x+1|<\delta\implies|f(x)-f(-1)|=|f(x)|<\epsilon$.

Note that

$$|f(x)|=\bigg|\frac{3x^{2}-2x-5}{2x+3}\bigg|=\bigg|\frac{(3x-5)(x+1)}{2x+3}\bigg|=\bigg|\frac{3x-5}{2x+3}\bigg||x+1|.$$

We are free top choose $\delta$ as we wish, and, in doing so, we will have a handle on making the $|x+1|$ term above as small as we want. To help us minimize $\big|\frac{3x-5}{2x+3}\big|$, let's see what happens if we would ensure $\delta <\frac{1}{3}$.

If $|x+1|<\frac{1}{3}$, then $-\frac{1}{3}<x+1<\frac{1}{3}$. This implies that

$$-1<3x+3<1\implies-9<3x-5<-7<9\implies|3x-5|<9$$

and

$$-\frac{2}{3}<2x+2<\frac{2}{3}\implies\frac{1}{3}<2x+3<\frac{5}{3}\implies\frac{1}{3}<|2x+3|.$$

Thus,

$$\bigg|\frac{3x-5}{2x+3}\bigg|<\frac{9}{1/3}=27.$$

So, if $\delta<\frac{1}{3}$, it follows that $|f(x)|<27|x+1|$. So, if in addition to this, $\delta<\frac{\epsilon}{27}$, we get that $|f(x)|<27\cdot\frac{\epsilon}{27}=\epsilon$.

Hence, we could choose $\delta=\min\{\frac{1}{3},\frac{\epsilon}{27}\}$. Then, if $|x+1|<\delta$, it follows that $|f(x)|<\epsilon$, as desired.

• Your choice of $1/3$ is arbitrary, am I right? – Miguelgondu May 14 '16 at 16:50
• almost. I needed any number $<1/2$. Otherwise I would get $-1/2<x+1<1/2\implies -1<2x+2<1\implies 0<2x+3<2$. And this is a problem since I'd be dividing by $0$. I need $2x+3>\eta>0$ for some $\eta$. – ervx May 14 '16 at 17:41
• Hey, thanks a lot. Just 1 more question, I thought that you'd need to choose a delta less than -3/2 so that the undefined point isn't in the deleted neighbour hood. How come you didn't do that? – Jacob May 15 '16 at 10:38
• Also, one more thing. Is there a reason why you chose 9 rather than -7 to be the max? Could you choose, say, 5 rather than the 9? – Jacob May 15 '16 at 11:23
• Since we are choosing $\delta<1/3$, we are only looking at the interval $(-4/3, -2/3)$, and the function is defined for all values in this range. The reason we needed $\delta<1/2$ was precisely so that we don't have to consider the value $x=-3/2$, because the function is unbounded as $x$ approaches $-3/2$. To answer the 2nd question, we do in fact need $9$, because we want to conclude that the absolute $|3x-5|<9$. To do this, we must show that $-9<3x-5<9$. If we just had $-9<3x-5<5$, then it's possible that $3x-5=-8$ and hence $|3x-5|=8>5$. – ervx May 15 '16 at 14:55

Fix $\varepsilon>0$ and let $x\in\mathbb{R}\setminus\{-\frac{3}{2}\}$ such that $|x-(-1)|=|x+1|<\delta$ for $\delta>0$ to be chosen later. As $f(-1)=0$, we have : \begin{align} \left|f(x)-f(-1)\right|&=\left|\frac{3x^2-2x-5}{2x+3}-0\right|\\ &=\left|\frac{3x^2-2x-5}{2x+3}\right|. \end{align} Now observe that \begin{align} 3x^2-2x-5&=3(x+1)\left(x-\frac{5}{3}\right),\quad\quad\quad\quad(*) \end{align} \begin{align} \left|x-\frac{5}{3}\right|&=\left|x+1-1-\frac{5}{3}\right|\\ &\leq\left|x+1\right|+\left|1+\frac{5}{3}\right|\\ &<\delta+\frac{8}{3},\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(**) \end{align} and that (by the inverse triangle inequality) \begin{align} |2x+3|&\geq|2|x|-3|\\ &\geq3-2|x|\\ &>3-2(1+\delta)\\ &=1-2\delta\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(***) \end{align} so that : \begin{align} \left|\frac{3x^2-2x-5}{2x+3}\right|&=\frac{3|x+1|\left|x-\frac{5}{3}\right|}{|2x+3|}\\ &<\frac{3\delta\left(\delta+\frac{8}{3}\right)}{|1-2\delta|}. \end{align} We now choose $\delta<1/4$ for $1-2\delta>1/2>0$, and then \begin{align} \left|f(x)-f(-1)\right|&<6\delta\left(\delta+\frac{8}{3}\right). \end{align} Finally, if we choose $\varepsilon=6\delta\left(\delta+\frac{8}{3}\right)$ for any $\delta<1/4$, then we get $$|x-(-1)|<\delta\quad\Longrightarrow\quad|f(x)-f(-1)|<\varepsilon$$ whence $f$ is continuous at $x=-1$.

This is perhaps not your assignment, but there are more general ways to apply the $\epsilon-\delta$ argument, in order to simplify problems like this.

Here are a few, although some of the proofs might gloss over a more carefully written proof:

$\bf{Lemma}$: The constant function $c: \mathbb{R} \to \mathbb{R}$ defined by $c(x)=k$ is everywhere continuous:

Let $\epsilon>0$, then let $\delta=\epsilon$. Clearly $|c(x)-c(y)|=0<\epsilon$.

$\bf{Lemma}$: The identity function $i: \mathbb{R} \to \mathbb{R}$ defined by $i(x)=x$ is everywhere continuous:

Let $\epsilon>0$, let $\delta=\epsilon$. Then suppose $|x-y|<\delta$. This implies that $|i(x)-i(y)|=|x-y|<\delta=\epsilon$.

$\bf{Theorem}$: Let $A \subseteq \mathbb{R}$ be a nonempty set, let $x \in A$, let $f,g:A \to \mathbb{R}$ be continuous and let $k \in \mathbb{R}$ Suppose that $f,g$ are continuous at $x$.

$\bf{1.}$ $f\pm g$ is continuous at $x$

$\bf{2.}$ $kf$ is continuous at $x$.

$\bf{3.}$ $fg$ is continuous at $x$.

$\bf{4.}$ if $g(x) \neq 0$, then $\frac{f}{g}$ is continuous at $x$.

proof: We show wlog that the result holds for $f+g$. Let $\epsilon>0$. Then there exist $\delta_f$ and $\delta_g$ so that $|x-y|<\delta_f \implies |f(x)-f(y)|<\epsilon/2$ and $|x-y|<\delta_g \implies |g(x)-g(y)|<\epsilon/2$. Let $\delta=\min\{\delta_f, \delta_g\}$. Suppose that $|x-y|<\delta$. Then, by the triangle inequality, we obtain: $$|(f+g)(x)-(f+g)(y)|=|f(x)-f(y)+g(x)-g(y)| \leq |f(x)-f(y)|+|g(x)-g(y)|<\epsilon/2+\epsilon/2=\epsilon$$

We now show that $kf$ is continuous at $x$.Suppose wlog that $k>0$. Let $\epsilon>0$. By assumption, there exists $\delta$ so that $|x-y|<\delta$ implies that $|f(x)-f(y)|<\frac{\epsilon}{k}$. Suppose that $|x-y|<\delta$. Then $$|kf(x)-kf(y)| =k \cdot |f(x)-f(y)|<k \cdot \frac{\epsilon}{k}=\epsilon.$$

*see if you can prove (3) and (4), they go very similarly.

$\bf{corollary \, 1}$: If $f,g: A \to \mathbb{R}$ are continuous, then the previous theorem holds for all of $A$.

$\bf{corollary \, 2}$: Polynomials $p: A \to \mathbb{R}$ are continuous.

$\bf{corollary \, 3}$: Rational functions $r: A \to \mathbb{R}$, defined by $r(x)=\frac{p_1(x)}{p_2(x)}$ for polynomials $p_1$ and $p_2$, are continuous at all $x \in A$ for which $p_2(x) \neq 0$.

Your problem is a special case of this.