1
$\begingroup$

I have the following problem:

A student wants to use the $(\epsilon , \delta)$-definition of limit to prove that $\lim_{x \to 1} f(x)=2$ for some function $f$. After analyzing the problem she finds that, for every $x \in \left(\frac{1}{2},\frac{3}{2}\right)$, $$|f(x)-2| \leq \frac{2|x-1|}{x^2}.$$ Given $\epsilon>0$, find a suitable $\delta>0$ that completes her proof.

Now, I'm fairly comfortable with simple Epsilon-Delta proofs, but I really don't know how to go about this problem. In the inequality, I know that the expressions inside absolute values resemble the $0<|x-a|<\delta$ and $|f(x)-L|<\epsilon$ parts of the definition, and I'm guessing that I have to use the fact that $x \in \left(\frac{1}{2},\frac{3}{2}\right)$ to find a suitable $\epsilon$, but that's about it; other than pushing symbols around to see if something makes sense, I don't have anything else.

Maybe there's something really obvious that I'm not seeing, but I'm stuck. I've looked for similar problems but I haven't found one similar to this one, so any help would be greatly appreciated.

$\endgroup$
3
$\begingroup$

We have $$|f(x)-2| \leq \frac{2|x-1|}{x^2}$$

Essentially, we need some sort of lower bound on $x^2$. For $x$ within the interval $(\frac 12, \frac 32)$, the lower bound $(\frac 14)^2 = \frac{1}{16}$ works(recall that $x^2$ is strictly increasing for $x>0$). Hence

$$|f(x)-2| \leq \frac{2|x-1|}{x^2} < 32|x-1| < \epsilon $$ and so we choose $\delta < \min({\frac 12, \frac{\epsilon}{32}})$

$\endgroup$
1
$\begingroup$

$$|f(x)-2|\leq \frac{2|x-1|}{x^2} \tag 1$$

Another approach. We know that $$x\mapsto x-1\\ x\mapsto|x|\\x\mapsto x^2\\x\mapsto 2x$$ are continuous functions, so the composition $$g:x\mapsto 2\frac {|x-1|}{x^2}$$

Is continuous, provided $x\neq 0$. That is $\lim_{x\to a}g(x)=g(a)$ for every $a\neq 0$.

In particular, it holds for $a=2$. So we know that for every $\varepsilon>0$ there exists some $\delta>0$ such that $|g(x)-g(a)|<\varepsilon$, provided $|x-a|<\delta$. As $g(a)=0$, we find our required $\delta$.


Let $\varepsilon>0$ be given. From above, there's a delta such that if $|x-1|<\delta$, then $\frac{2|x-1|}{x^2}<\varepsilon$. Using $(1)$ above, we get

$$ |f(x)-2|\leq \frac{2|x-1|}{x^2}<\varepsilon $$

i.e, $|f(x)-2|<\varepsilon$ provided $x\in (2-\delta,2+\delta)$, and we're done.

$\endgroup$
2
  • $\begingroup$ So, correct me if I'm wrong, but basically because of the Squeeze Theorem whatever $\delta$ that works for $g(x)$ will also work for $f(x)$, right? $\endgroup$
    – Eduardo M.
    May 7 '16 at 22:25
  • $\begingroup$ @EduardoM. I've added a few more details, there's no need to invoke the Squeeze theorem. $\endgroup$ May 7 '16 at 22:33

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.