# How do you prove the statement lim as x approaches 9 for root4(9-x) = 0 using the epsilon and delta definition?

Dec 19, 2016

See explanation.

#### Explanation:

The $\delta \text{-} \epsilon$ definition of a limit states that:

the limit of a function $f \left(x\right)$, as $x$ approaches some value $c$, is $L$ if, for every possible $\epsilon > 0$, we can find a $\delta > 0$ that depends on $\epsilon$, such that $\left\mid f \left(x\right) - L \right\mid < \epsilon$ whenever $\left\mid x - c \right\mid < \delta$.

It's like a game. Player 1 picks an $\epsilon > 0$, and Player 2 is trying to find a $\delta > 0$ such that every $x$ within $\pm \delta$ of $c$ (that is, $\textcolor{n a v y}{x \in \left(c - \delta , c + \delta\right)}$) is guaranteed to get mapped to an $f \left(x\right)$ within $\pm \epsilon$ of $L$ (that is, $\textcolor{g r e e n}{f \left(x\right) \in \left(L - \epsilon , L + \epsilon\right)}$). Player 1 keeps picking smaller and smaller $\epsilon$, and Player 2 keeps having to find smaller and smaller $\delta$.

If we can prove that, for every $\epsilon$ Player 1 picks, Player 2 can find a suitable $\delta$, then we've proven the limit. In other words:

If

$\forall \text{ "epsilon > 0" "EE" "delta>0}$

such that

$\left\mid f \left(x\right) - L \right\mid < \epsilon$ when $\left\mid x - c \right\mid < \delta$

then

${\lim}_{x \to c} f \left(x\right) = L$.

Okay, so that's a lot of mumble jumble. Let's put it to use.

We need to find a $\delta$ that depends on $\epsilon$. That means you can think of $\delta$ as a function of $\epsilon$. We also want to have $' x \text{ is within "delta" of } c '$ imply $' f \left(x\right) \text{ is within " epsilon " of } L '$, or in math lingo:

$\left\mid x - c \right\mid < \delta \implies \left\mid f \left(x\right) - L \right\mid < \epsilon$

So we start with $\left\mid x - c \right\mid < \delta$ and $\textcolor{b l u e}{\text{try to get it to look like } \left\mid f \left(x\right) - L \right\mid < \epsilon}$.

In this example, $c = 9$, $L = 0$, and $f \left(x\right) = \sqrt{9 - x}$.

$\left\mid x - c \right\mid < \delta \implies \left\mid x - 9 \right\mid < \delta$
$\textcolor{w h i t e}{\left\mid x - c \right\mid < \delta} \implies \left\mid 9 - x \right\mid < \delta$
$\textcolor{w h i t e}{\left\mid x - c \right\mid < \delta} \implies \sqrt{\left\mid 9 - x \right\mid} < \sqrt{\delta}$
$\textcolor{w h i t e}{\left\mid x - c \right\mid < \delta} \implies \left\mid \sqrt{9 - x} \right\mid < \sqrt{\delta}$
$\textcolor{w h i t e}{\left\mid x - c \right\mid < \delta} \implies \left\mid \sqrt{9 - x} - 0 \right\mid < \sqrt{\delta}$
$\textcolor{w h i t e}{\left\mid x - c \right\mid < \delta} \implies \left\mid f \left(x\right) - 0 \right\mid < \sqrt{\delta}$

Hey—looks like we may have found our connection between $\delta$ and $\epsilon$! If we let $\sqrt{\delta} = \epsilon$, then we have

$\textcolor{w h i t e}{\left\mid x - c \right\mid < \delta} \implies \left\mid f \left(x\right) - 0 \right\mid < \epsilon$

and so we've shown that $\left\mid x - 9 \right\mid < \delta \implies \left\mid f \left(x\right) - 0 \right\mid < \epsilon$.

The last thing to do is to solve $\sqrt{\delta} = \epsilon$ for $\delta$:

$\text{ } \sqrt{\delta} = \epsilon$
$\implies \delta = {\epsilon}^{4}$

All the work we've done here simply means that no matter how small an $\epsilon$ Player 1 may pick, Player 2 can always just choose their $\delta$ to be ${\epsilon}^{4}$, and they'll win every time. That is, as long as $x$ is within ${\epsilon}^{4}$ of $9$, $\sqrt{9 - x}$ will be within $\epsilon$ of $0$. Thus,

${\lim}_{x \to 9} \sqrt{9 - x} = 0$

has been proven.

$Q E D .$