Show that $\lim_{x\to 4} \sqrt{x} = 2$ using $\epsilon$-$\delta$ definition of limit.

Question

Show that $\lim_{x\to 4} \sqrt{x} = 2$ using $\epsilon$-$\delta$ definition of limit.

Method 1

Let $\epsilon>0$

Choose $\delta<\epsilon-6$. ($\delta>0$)

If $|x-4|<\delta \implies -\delta < x-4 < \delta \implies 4-\delta<x<4+\delta$,

then, $|\sqrt{x}-2| < \sqrt{x}+2 < \sqrt{4+\delta}+2 < 6+\delta < \epsilon$

Hence, by $\epsilon$-$\delta$ definition of limit, $\lim_{x\to 4} \sqrt{x} = 2$

Is method 1 correct?

Method 2

Let $\epsilon>0$

Choose $\delta<$___. ($\delta>0$)

If $|x-4|<\delta \implies -\delta < x-4 < \delta \implies 4-\delta<x<4+\delta$,

then, $|\sqrt{x}-2| < \sqrt{x}+2 < \sqrt{4+\delta}+2$

(Scratch work: $\sqrt{4+\delta}+2=\epsilon \implies \delta+4=(\epsilon-2)^{2} \implies \delta=\epsilon^{2}-4\epsilon$.)

Now I'm tempted to fill the blank by choosing $\delta<\epsilon^{2}-4\epsilon$. But when I do that I'm not getting the desired $|\sqrt{x}-2| < \epsilon$ because $\sqrt{4+\delta}+2<\sqrt{4+\epsilon^{2}-4\epsilon}+2=\sqrt{(\epsilon-2)^{2}}+2=|\epsilon-2|+2$, which is not necessarily less than $\epsilon$.

What's wrong in method 2?

Answer 1

Scratch Work

The goal is to determine a value of $\delta > 0$ such that $|x-4| < \delta$ implies that $|\sqrt{x} - 2| < \varepsilon$. Note

$$ \bigl| x-4 \bigr| = \bigl| (\sqrt{x}-2)(\sqrt{x}+2) \bigr|. \label{eq1}\tag{*}$$

Observe that $| x-4 | < \delta$ implies \begin{align} -\delta < x-4 < \delta &\iff 4-\delta < x < 4+\delta \\ &\iff \sqrt{4-\delta} < \sqrt{x} < \sqrt{4+\delta} \\ &\iff \sqrt{4-\delta}+2 < \sqrt{x}+2 < \sqrt{4+\delta}+2. \end{align} Hence $\sqrt{x}+2$ it is possible to conclude that $\sqrt{x}+2 > 2$ for any value of $\delta$ such that $\sqrt{4-\delta}$ "makes sense", i.e. $\delta < 4$. This provides some control over the second factor in (\ref{eq1}) (it will be bigger than $1$, which is helpful). The other factor is $\sqrt{x}-2$, which is the thing we are trying to make small. In particular, $$\bigl| \sqrt{x}-2 \bigr| < \bigl| (\sqrt{x}-2)(\sqrt{x}+2) \bigr| = \bigl| x-4\bigr| < \delta. $$ Of particular note is that $|\sqrt{x}-2|$ is now bounded by $\delta$, which is exactly what is desired, as long as $\delta < \varepsilon$.

So choose $\delta$ to be smaller than $4$ (to bound $\sqrt{x}+2$), and smaller than $\varepsilon$ (to get the desired result in the end).

Proof

Fix $\varepsilon > 0$ and choose $\delta < \min\{4, \varepsilon\}$. Suppose that $|x-4| < \delta$. Since $\delta < 4$, if $|x-4|<\delta$, then $$ |x-4| < 4 \implies 0 < x \implies 2 < \sqrt{x}+2. $$ In particular, $\sqrt{x}+2 > 1$, hence \begin{align} \bigl|\sqrt{x} - 2\bigr| &< \bigl| (\sqrt{x}+2)(\sqrt{x}-2)\bigr| && \text{(since $\sqrt{x}+2 > 1$)} \\ &= \bigl| x-4 \bigr| \\ &< \delta \\ &< \varepsilon, \end{align} which is the desired result.

Answer 2

As noted in the comments, both your methods are wrong due to a wrong choice of $\delta$. One thing I find helpful when trying to proof these kind of limits is to choose multiple $\delta$ for different cases and then take the minimal $\delta$. So in your example, take $|\sqrt{x}-2|$. You have shown already, that $4-\delta<x<4+\delta$ for any choice of $\delta$. There are two cases to consider:

1.$|\sqrt{x}-2|=\sqrt{x}-2<\sqrt{(\epsilon+2)^2}-2=\epsilon$, as $\epsilon>0$, so $\epsilon+2>0$ with $\delta = (\epsilon+2)^2-4(>0)$.
2.$|\sqrt{x}-2|=2-\sqrt{x}<2-\sqrt{4-\delta}$ If you assume $0<4-\delta$, which you are allowed to. Now you can calculate from $2-\sqrt{4-\delta}\leq\epsilon$ that $0<4-(2-\epsilon)^2\leq\delta$.
By taking $\delta_m=\min\{((\epsilon+2)^2-4,4-(2-\epsilon)^2,4\}$ you get all the desired inequalities and have proven your limit.

Answer 3

HINT

Another possible approach. Let $|x - 4| < \delta_{\varepsilon}$. Then one concludes that: \begin{align*} |\sqrt{x} - 2| = \frac{|x - 4|}{\sqrt{x} + 2} < \frac{|x - 4|}{2} < \frac{\delta_{\varepsilon}}{2} < \varepsilon \end{align*}

Can you proceed from here?