Given $f(x)= {\sqrt x}$, L = 2, $\epsilon = {\frac 1 5}$ and C=4.

I know that I start with $$|{\sqrt x} -2| \le {\frac 1 5}$$

If I rewrite that as $$-{\frac 1 5} \le {\sqrt x} -2 \le {\frac 1 5}$$

I can then add 2 to both sides and square everything to get rid of the radical which would give me

$$\frac{81}{25} \lt x \lt \frac{121}{25}$$

but something doesn't seem right with that approach.

  • $\begingroup$ Please clarify: The question is not only (1) Prove formally that $\lim_{x \to 4}\sqrt{x} = 2$, but also (2) Find the largest possible $\delta\gt{0}$ such that $|x-4|\lt\delta\implies|\sqrt{x}-2|\lt\frac{1}{5}$? $\endgroup$ – user695163 Aug 28 '19 at 4:16
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    $\begingroup$ In this case it was only to find the largest possible $\delta$ $\endgroup$ – dstarh Aug 28 '19 at 4:28

Nothing wrong with the approach so far, aside from mixing up $<$ with $\le$ sometimes. What you have shown so far is $$\frac{81}{25} < x < \frac{121}{25} \iff |\sqrt{x} - 2| < \frac{1}{5}.$$ Further, we can say, $$-\frac{19}{25} < x - 4 < \frac{21}{25} \iff |\sqrt{x} - 2| < \frac{1}{5}.$$ Our largest possible value of $\delta$ is therefore $\frac{19}{25}$; any larger value of $\delta$ will allow for $x - 4$ to be too large and negative. We now have \begin{align*} |x - 4| < \frac{19}{25} &\iff -\frac{19}{25} < x - 4 < \frac{19}{25} \\ &\implies -\frac{19}{25} < x - 4 < \frac{21}{25} \\ &\iff |\sqrt{x} - 2| < \frac{1}{5}, \end{align*} completing the proof for this particular $\varepsilon$.

So, why aren't $\varepsilon$-$\delta$ proofs this straightforward, generally speaking? Well, the idea here is to examine the condition $|f(x) - L| < \varepsilon$, and figure out the set of $x$ for which this condition holds true. In our case, we found it was the open interval $\left(\frac{81}{25}, \frac{121}{25}\right)$. Most of the time, this method turns out to either be either intractable, or at least, a far more difficult way of going about things.

Consider, for example, proving $f(x) = x^4 + 2x$ is continuous at $x = 1$. If we fix $\varepsilon = \frac{1}{5}$ as before, then now we need to solve the inequality: $$-\frac{1}{5} < x^4 + 2x - 3 < \frac{1}{5}$$ Solving this involves finding the roots of the two quartics $x^4 + 2x - 3 \pm \frac{1}{5}$, a problem that is possible to do, but horribly, horribly complicated. If we consider higher degree polynomials, it can easily be impossible to do with exact values.

It's much easier instead to make estimates. You don't need to find all $x$ such that $|f(x) - L| < \varepsilon$, but just an interval an interval around $x = c$. This can yield much more straightforward proofs that are easier to read, write, and come up with!

  • $\begingroup$ Thanks, how did you go from -19/25 to 19/25? $\endgroup$ – dstarh Aug 28 '19 at 4:11
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    $\begingroup$ $19/25$ is the largest (absolute) distance you can go from $4$ before you step outside the range of values for $x$. $\endgroup$ – Theo Bendit Aug 28 '19 at 4:12

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