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.


2 Answers 2


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 $\delta < \min({\frac 12, \frac{\epsilon}{32}})$


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

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.

  • $\begingroup$ 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, 2016 at 22:25
  • $\begingroup$ @EduardoM. I've added a few more details, there's no need to invoke the Squeeze theorem. $\endgroup$ May 7, 2016 at 22:33

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