# Epsilon-delta continuity proof

#### GreenGoblin

##### Member
Show that the following are continuous at x=1 using the epsilon-delta definition:

$x^{2} - x + 1$

$\sqrt (x)$

I know the definitions but I don't really know quite what to do with them. After the simple rearranging I'm just at a bit of a dead end; any pointers?

Gracias,
GreenGoblin

Last edited:

#### Prove It

##### Well-known member
MHB Math Helper
Show that the following are continuous at x=1 using the epsilon-delta definition:

$x^{2} - x + 1$

$\sqrt (x)$

I know the definitions but I don't really know quite what to do with them. After the simple rearranging I'm just at a bit of a dead end; any pointers?

Gracias,
GreenGoblin
For a function to be continuous at a point, a limit needs to exist at that point.

By definition, if $\displaystyle 0 < |x - c| < \delta \implies \left|f(x) - L\right| < \epsilon$, then $\displaystyle \lim_{x \to c}f(x) = L$

So for the first one, to show that $\displaystyle \lim_{x \to 1}\left(x^2 - x + 1\right) = 1$, we need to show that $\displaystyle 0 < |x - 1| < \delta \implies \left|\left(x^2 - x + 1\right) - 1\right| < \epsilon$.

Trying to solve $\left|\left(x^2 - x + 1\right) - 1\right| < \epsilon$ for $\displaystyle |x - 1|$ gives us...

\displaystyle \begin{align*} \left|x^2 - x\right| &< \epsilon \\ \left|x\left(x - 1\right)\right| &< \epsilon \\ |x||x - 1| &< \epsilon \end{align*}

Now define $\displaystyle M$ so that $\displaystyle |x| < M$ and then we have

\displaystyle \begin{align*} M|x - 1| &< \epsilon \\ |x - 1| &< \frac{\epsilon}{M} \end{align*}

Suppose that we make $\displaystyle |x - 1| < \frac{1}{2}$, in other words, ensuring that the distance from x and 1 is never any more than 1/2 a unit (letting $\displaystyle \delta = \frac{1}{2}$, which we can do because we are going to close in on x = 1 by making that distance small anyway), and we find...

\displaystyle \begin{align*} |x - 1| &< \frac{1}{2} \\ -\frac{1}{2} < x - 1 &< \frac{1}{2} \\ \frac{1}{2} < x &< \frac{3}{2} \end{align*}

and therefore we can let $\displaystyle M = \frac{3}{2}$, so

\displaystyle \begin{align*} |x - 1| &< \frac{\epsilon}{\frac{3}{2}} \\ |x - 1| &< \frac{2}{3}\epsilon \end{align*}

Therefore, we can define $\displaystyle \delta = \min\left\{ \frac{1}{2}, \frac{2}{3}\epsilon \right\}$ and reverse each step, and you will have your proof

#### Prove It

##### Well-known member
MHB Math Helper
Show that the following are continuous at x=1 using the epsilon-delta definition:

$x^{2} - x + 1$

$\sqrt (x)$

I know the definitions but I don't really know quite what to do with them. After the simple rearranging I'm just at a bit of a dead end; any pointers?

Gracias,
GreenGoblin
As for the second, we would need to show that \displaystyle \begin{align*} 0 < |x - 1| < \delta \implies \left| \sqrt{x} - 1 \right| < \epsilon \end{align*}.

Solving the second inequality for the first gives us...

\displaystyle \begin{align*} \left| \sqrt{x} - 1 \right| &< \epsilon \\ \sqrt{x} - 1 &< \epsilon ^2 \\ \sqrt{x} &< 1 + \epsilon ^2 \\ x &< \left( 1 + \epsilon ^2 \right)^2 \\ x &< 1 + 2\epsilon ^2 + \epsilon ^4 \\ x - 1 &< 2\epsilon ^2 + \epsilon ^4 \\ |x - 1 | &< 2\epsilon ^2 + \epsilon ^4 \textrm{ which we can do because this is a nonnegative quantity} \end{align*}

So if we let \displaystyle \begin{align*} \delta = 2 \epsilon ^2 + \epsilon ^4 \end{align*} and reverse the process, we will have our proof

#### Deveno

##### Well-known member
MHB Math Scholar
a general word on how these types of proofs go:

you assume that $\epsilon$ is given before-hand. it might be arbitrarily small, though, so you're going to need to a pretty small $\delta$ to make it work. how small?

if you can express $\delta$ in terms of $\epsilon$ that's OK.

but normally, you have something like $|x - a| < \delta$ as the condition $x$ must satisfy to get $|f(x) - f(a)| < \epsilon$.

$|f(x) - f(a)| < \epsilon$

you want to re-arrange that (using every/any algebraic trick you can think of) to

$|x - a|\text{..something...} = \text{...expression involving } \epsilon$

some tips:

you can always require that $|x - a| < M$ and take the smaller of your formula involving $\epsilon$ and M to be $\delta$...often this makes some "mess" go away.

the triangle inequality is useful for splitting up sums inside an absolute value sign.

you can always factor products outside the absolute value sign $|ab| = |a|\cdot |b|$.

it's a good idea to have a mental picture of where "a" is: for example, if a > 0, then by choosing M small enough (as above in tip #1), sometimes we can "lose the absolute value signs", which gives us a bit more freedom with the algebra.

the best possible situation to have is finagling $|f(x) - f(a)| < \epsilon$ into something like:

$|x - a|\cdot|K| < g(\epsilon)$ where the expression $g$ does NOT have $x$ in it (important!). $K$ might be a very ugly-looking formula involving $a$, but that doesn't matter, as long as it is either:

1) constant, or
2) bounded by a positive constant

situation (2) will occur more often. that is where choosing a good $M$ will come in handy.