2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is there any reason in particular that this example is difficult/interesting for you? $\endgroup$ – Andres Mejia May 14 '16 at 15:14
  • 1
    $\begingroup$ 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 $\epsilon$" $\endgroup$ – H. Potter May 14 '16 at 15:16
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your choice of $1/3$ is arbitrary, am I right? $\endgroup$ – Miguelgondu May 14 '16 at 16:50
  • 1
    $\begingroup$ 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 $\eta$. $\endgroup$ – ervx May 14 '16 at 17:41
  • $\begingroup$ 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? $\endgroup$ – Jacob May 15 '16 at 10:38
  • $\begingroup$ 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? $\endgroup$ – Jacob May 15 '16 at 11:23
  • $\begingroup$ 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 $|3x-5|=8>5$. $\endgroup$ – ervx May 15 '16 at 14:55
2
$\begingroup$

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 $x=-1$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.