Verify the Limit: $\lim_{x\rightarrow 9} (\sqrt{x} +9) = 12$

Question

I am using the definition of the limit of a function. So far I have:

$$|f(x)-L|<\epsilon \iff |(\sqrt{x}+9)-12|<\epsilon \iff |\sqrt{x}-3|<\epsilon.$$

This is where I get stuck. Any help is appreciated.

Answer 1

Keep in mind that in deriving these proofs we are always working backwards. We start with $\epsilon$ and try to find $\delta$ that will make it true.

Let $\epsilon > 1$. Since $\epsilon $ may be as small as we like we may as well assume $1 > \epsilon > 0$. I'm saying this now, because later we will have a reason for this.

So we want to find a $\delta$ (expressed in terms of $\epsilon$) so that whenever $|x - 9|< \delta$ we will have to have $|\sqrt{x} - 3|< \epsilon$.

We can find this $\delta$ one of two ways:

Forward:

$|x - 9|<\delta \implies$

$-\delta < x - 9 < \delta \implies $

$9 -\delta < x < 9+\delta \implies $

$\sqrt{9-\delta} < x <\sqrt{9+\delta}\implies$

Now if we can somehow find some way that $3 - \epsilon \le \sqrt{9-\delta}$ and then $\sqrt{9+\delta} \le 3+ \epsilon$ we'll be on our way.

So we want $3-\epsilon \le \sqrt{9 -\delta}$ so $9 - 6\epsilon + \epsilon^2 \le 9- \delta$. So we need $\delta \le 6\epsilon - \epsilon^2$.

And we want $\sqrt {9-\delta} \le 3 + \epsilon$ so $9-\delta \le 9 + 6\epsilon +\epsilon^2$. So we need $-6\epsilon -\epsilon^2 \le \delta$. Well, that's true for any positive $\delta$.

How can we find a $\delta$ where both will be true. Well, if we assume $0 < \epsilon < 1$, then $\epsilon^2 < \epsilon$ so if we can define $\delta$ so that $\delta \le 5\epsilon < 6\epsilon - \epsilon^2$ we'd be done.

So define $\delta = 5\epsilon$. If $|x - 9| < \delta$ then $-5\epsilon < x-9 < 5\epsilon$ and $9-6\epsilon + \epsilon^2 < 9-5\epsilon < x < 9+5\delta < 9 + 6\delta + \delta^2$. So $ 3-\epsilon < \sqrt {9-5\epsilon} < \sqrt x< \sqrt{9+5\delta} < 3 + \epsilon$. So $-\epsilon < \sqrt x < \epsilon$ and $|\sqrt x - 3| < \epsilon$.

The proof is done.

Or backwards:

$|\sqrt{x} - 3| <\epsilon $ is implied by

$-\epsilon < \sqrt{x} - 3 <\epsilon \Leftarrow$

$3-\epsilon < \sqrt{x} < 3 + \epsilon\Leftarrow$

$9-6\epsilon +\epsilon^2 < x < 9 + 6\epsilon + \epsilon^2 \Leftarrow$

$-6\epsilon +\epsilon^2 < x -9 < 6\epsilon + \epsilon^2 \Leftarrow$

$-5\epsilon < -6\epsilon +\epsilon^2 < x-9 < 5\epsilon < 6\epsilon + \epsilon^2$.

Note: $\Leftarrow$ means is implied by. We are going in the opposite direction then we are used to.

Normally we say $x -9 < 5\epsilon \implies x-9 < 5\epsilon < 6\epsilon + \epsilon^2\implies x< 6\epsilon + \epsilon^2$ because we are saying: if A then B.

But $\Leftarrow$ means: A can come about from B. Or: if B then A. $x-9 < 5\epsilon \Leftarrow x-9 < 6\epsilon + \epsilon^2$.

$\Leftarrow |x-9|< 5\epsilon$.

S if we let $\delta = 5\epsilon$ we are done.

So define $\delta = 5\epsilon$. If $|x - 9| < \delta$ then $-5\epsilon < x-9 < 5\epsilon$ and $9-6\epsilon + \epsilon^2 < 9-5\epsilon < x < 9+5\delta < 9 + 6\delta + \delta^2$. So $ 3-\epsilon < \sqrt {9-5\epsilon} < \sqrt x< \sqrt{9+5\delta} < 3 + \epsilon$. So $-\epsilon < \sqrt x < \epsilon$ and $|\sqrt x - 3| < \epsilon$.

The proof is done.

Answer 2

We want to show that for every $\epsilon>0$ there exists $\delta$ such that for all $x\in\mathbb{R}$, if $0<|x-9|<\delta$ then $$|\sqrt x-3|<\epsilon.$$ Assuming the principal root, we get $$|\sqrt x-3|\le|x-9|$$ so taking $\delta=\epsilon$ we have the required result.

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Can you algebraically check that the inequality in blue holds?

Answer 3

For the given $\epsilon,$ you need to find a $\delta $ such that $$0<|x-9|<\delta \implies |\sqrt x -3 |<\epsilon .$$

Let $$\delta = \epsilon $$

If $|x-9|<\delta$, we have

$$ |\sqrt x -3 |= \frac {|x-9|}{|\sqrt x +3 |} <|x-9| <\delta = \epsilon .$$