MHB Is the Function \( f(x) = \sqrt{4 + x^2} \) Continuous Everywhere?

[mathmari]

Hey!

I want to show that the function $f:\mathbb{R}\rightarrow \mathbb{R}$ with $f(x)=\sqrt{4+x^2}$ is continuous on $\mathbb{R}$ using the $\epsilon$, $\delta$-definition.

We have the following:

To show the continuity of $f$ we have to prove the continuity at each point $\displaystyle x_{0}\in \mathbb {R}$.
Let $x_{0}$ be an arbitrary real number.
We consider an arbitrary $\epsilon >0$.
We have to find a small enough $ \delta >0$ such that $|f(x)-f(x_{0})|<\epsilon$ for all $x$ with $|x-x_{0}|<\delta$.

First we are looking at the inequality $|f(x)-f(x_{0})|<\epsilon$ :
\begin{align*}&|f(x)-f(x_0)|<\epsilon \\ &\Rightarrow |\sqrt{4+x^2}-\sqrt{4+x_0^2}|<\epsilon\\ & \Rightarrow |\sqrt{4+x^2}-\sqrt{4+x_0^2}|\cdot \frac{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}<\epsilon \\ & \Rightarrow \frac{|(4+x^2)-(4+x_0^2)|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}<\epsilon \\ & \Rightarrow \frac{|x^2-x_0^2|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}<\epsilon \\ & \Rightarrow \frac{|x-x_0||x+x_0|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}<\epsilon\end{align*}

It must hold that $\frac{|x-x_0||x+x_0|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}<\epsilon$ for all $x$ with $|x-x_{0}|<\delta$.

So, we have to choose $\delta$ in such a way that from $|x-x_{0}|<\delta$ we get the inequality $\frac{|x-x_0||x+x_0|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}<\epsilon$.

We have that \begin{equation*}\frac{|x-x_0||x+x_0|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}\ \overset{|x-x_{0}|<\delta}{ < } \ \frac{\delta\cdot |x+x_0|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}\end{equation*}

How could we choose $\delta$ ? (Wondering)

 

[castor28]

Hi mathmari,

As we have $\sqrt{4+x^2}+\sqrt{4+x_0^2}\ge 4$, it is enough to choose $\delta$ such that $\left|x^2-x_0^2\right|<4\epsilon$; this is what you would do to prove that $x\mapsto x^2$ is continuous.

 

[mathmari]

As we have $\sqrt{4+x^2}+\sqrt{4+x_0^2}\ge 4$, it is enough to choose $\delta$ such that $\left|x^2-x_0^2\right|<4\epsilon$; this is what you would do to prove that $x\mapsto x^2$ is continuous.

We have that $|\sqrt{4+x^2}+\sqrt{4+x_0^2}|>|\sqrt{4}+\sqrt{4}|=2+2=4$.

So, we get that $$\frac{|x-x_0||x+x_0|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}<|x-x_0|\cdot \frac{|x+x_0|}{4}\leq |x-x_0|\cdot \frac{|x|+|x_0|}{4}$$

Since $|x-x_{0}|<\delta$, we have that $$-\delta<x-x_{0}<\delta\Rightarrow x_0-\delta<x<\delta+x_0$$ Do we get from that $$|x|<|\delta+x_0|<|\delta|+|x_0|$$ ?

If this is correct, then we get the following:
\begin{align*}\frac{|x-x_0||x+x_0|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}&< |x-x_0|\cdot \frac{|x|+|x_0|}{4} \\ & < \delta\cdot \frac{\delta+|x_0|+|x_0|}{4} \\ & =\delta\cdot \frac{\delta+2|x_0|}{4} \\ & =\frac{\delta^2+2\delta|x_0|}{4}\\
& <\frac{\delta^2+2\delta|x_0|+|x_0|^2}{4} \\ & =\frac{(\delta+|x_0|)^2}{4}\end{align*}

We want that $$\frac{(\delta+|x_0|)^2}{4}<\epsilon\Rightarrow (\delta+|x_0|)^2<4\epsilon \Rightarrow \delta+|x_0|<2\sqrt{\epsilon}\Rightarrow \delta<2\sqrt{\epsilon}+|x_0|$$

Is everything correct so far?

So, do we choose $\delta:=2\sqrt{\epsilon}+|x_0|$ ?

(Wondering)

 

[Opalg]

We want that $$\frac{(\delta+|x_0|)^2}{4}<\epsilon\Rightarrow (\delta+|x_0|)^2<4\epsilon \Rightarrow \delta+|x_0|<2\sqrt{\epsilon}\Rightarrow \delta<2\sqrt{\epsilon}+|x_0|$$

Is everything correct so far?

It's correct up until the last line. But if $ \delta+|x_0|<2\sqrt{\epsilon}$ then $\delta<2\sqrt{\epsilon} \mathbin{\color{red}-} |x_0|$. That unfortunately ruins the calculation, because $2\sqrt{\epsilon}-|x_0|$ might be negative, and we have to choose $\delta$ to be less than some positive quantity.

I would go back to this step:

If this is correct, then we get the following:
$$\begin{aligned}\frac{|x-x_0||x+x_0|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}&< |x-x_0|\cdot \frac{|x|+|x_0|}{4} \\ & < \delta\cdot \frac{\delta+|x_0|+|x_0|}{4} \\ & =\delta\cdot \frac{\delta+2|x_0|}{4}\end{aligned}$$

The trick then is to choose $\delta$ so that both factors in that product $\delta\cdot \frac{\delta+2|x_0|}{4}$ are sufficiently small. For example, you could start by making $\delta<1$. Then $\dfrac{\delta+2|x_0|}{4} < \dfrac{1+2|x_0|}{4}$, and so $\delta\cdot \dfrac{\delta+2|x_0|}{4} < \dfrac{\delta(1+2|x_0|)}{4}$. That will be less than $\epsilon$ provided that $$\delta < \frac{4\epsilon}{1+2|x_0|}.$$

To ensure that both those conditions on $\delta$ are satisfied, you need to take $\delta < \min\bigl\{1,\frac{4\epsilon}{1+2|x_0|}\bigr\}.$

 

[mathmari]

The trick then is to choose $\delta$ so that both factors in that product $\delta\cdot \frac{\delta+2|x_0|}{4}$ are sufficiently small. For example, you could start by making $\delta<1$. Then $\dfrac{\delta+2|x_0|}{4} < \dfrac{1+2|x_0|}{4}$, and so $\delta\cdot \dfrac{\delta+2|x_0|}{4} < \dfrac{\delta(1+2|x_0|)}{4}$. That will be less than $\epsilon$ provided that $$\delta < \frac{4\epsilon}{1+2|x_0|}.$$

To ensure that both those conditions on $\delta$ are satisfied, you need to take $\delta < \min\bigl\{1,\frac{4\epsilon}{1+2|x_0|}\bigr\}.$

Why can we take $\delta<1$ ?

So, we don't take a specific $\delta$ we just say that it must be $\delta < \min\bigl\{1,\frac{4\epsilon}{1+2|x_0|}\bigr\}$, do we?

(Wondering)

 

[I like Serena]

Why can we take $\delta<1$ ?

So, we don't take a specific $\delta$ we just say that it must be $\delta < \min\bigl\{1,\frac{4\epsilon}{1+2|x_0|}\bigr\}$, do we?

We can make $\delta$ as small as we want, meaning we can set any arbitrary upper bound on it.
That is because what we really want, is to find a $\delta > 0$ as function $\delta=\delta(\epsilon; x_0)$ of $\epsilon >0$ and $x_0$ such that the conditions are satisfied.
And if we need to, we can make always pick a function that is smaller by a factor, or an upper bound, or whatnot. (Nerd)In our case, as Opalg explained, we can pick $\delta=\delta(\epsilon; x_0) = \min\bigl\{1,\frac{4\epsilon}{1+2|x_0|}\bigr\}$.
Or alternatively $\delta=\delta(\epsilon; x_0) = \min(1,4\epsilon)$, which is smaller.
Or alternatively we just say that we pick a $\delta=\delta(\epsilon; x_0)$ that is just smaller than a given expression. And if we want to make it explicit, we can just multiply the given expression by $\frac 12$.In case of doubt, we should check if $|f(x)-f(x_0)| < \epsilon$ does indeed hold for any $x$ with $0<|x-x_0|<\delta(\epsilon; x_0)$. (Sweating)

 

[mathmari]

We can make $\delta$ as small as we want, meaning we can set any arbitrary upper bound on it.
That is because what we really want, is to find a $\delta > 0$ as function $\delta=\delta(\epsilon; x_0)$ of $\epsilon >0$ and $x_0$ such that the conditions are satisfied.
And if we need to, we can make always pick a function that is smaller by a factor, or an upper bound, or whatnot. (Nerd)In our case, as Opalg explained, we can pick $\delta=\delta(\epsilon; x_0) = \min\bigl\{1,\frac{4\epsilon}{1+2|x_0|}\bigr\}$.
Or alternatively $\delta=\delta(\epsilon; x_0) = \min(1,4\epsilon)$, which is smaller.
Or alternatively we just say that we pick a $\delta=\delta(\epsilon; x_0)$ that is just smaller than a given expression. And if we want to make it explicit, we can just multiply the given expression by $\frac 12$.

Ah ok!

In case of doubt, we should check if $|f(x)-f(x_0)| < \epsilon$ does indeed hold for any $x$ with $0<|x-x_0|<\delta(\epsilon; x_0)$. (Sweating)

Let ${\displaystyle \epsilon >0}$ and let $\delta < \min\bigl\{1,\frac{4\epsilon}{1+2|x_0|}\bigr\}$. If ${\displaystyle \left|x-x_{0}\right|<\delta }$ then we get the following: :
\begin{align*}|f(x)-f(x_0)|&= |\sqrt{4+x^2}-\sqrt{4+x_0^2}|\\ & = |\sqrt{4+x^2}-\sqrt{4+x_0^2}|\cdot \frac{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|} \\ & =\frac{|(4+x^2)-(4+x_0^2)|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|} \\ & = \frac{|x^2-x_0^2|}{|\sqrt{4+x^2}+\sqrt{4+x_0^2}|} \\ & < \frac{|x^2-x_0^2|}{4} \\ & = \frac{|x-x_0||x+x_0|}{4} \\ & < \frac{\delta\cdot|x+x_0|}{4} \\ & = \frac{\delta\cdot|x-x_0+x_0+x_0|}{4} \\ & = \frac{\delta\cdot|(x-x_0)+2x_0|}{4} \\ & \leq \frac{\delta\cdot (|x-x_0|+|2x_0|)}{4} \\ & < \frac{\delta\cdot (\delta+|2x_0|)}{4} \\ & \overset{ \delta\leq 1 }{ \leq } \frac{\delta\cdot (1+|2x_0|)}{4} \\ & \overset{ \delta\leq \frac{4\epsilon}{1+2|x_0|} }{ \leq } \frac{4\epsilon}{1+2|x_0|}\cdot \frac{ (1+|2x_0|)}{4}\\ & = \epsilon\end{align*}
right? (Wondering)

About Lipschitz-continuity:

Let $x,y\in [a,b]$ then we have that
\begin{align*}|f(x)-f(y)|&= |\sqrt{4+x^2}-\sqrt{4+y^2}|\\ & = |\sqrt{4+x^2}-\sqrt{4+y^2}|\cdot \frac{|\sqrt{4+x^2}+\sqrt{4+y^2}|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|} \\ & =\frac{|(4+x^2)-(4+y^2)|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|} \\ & = \frac{|x^2-y^2|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|} \\ & < \frac{|x^2-y^2|}{4} \\ & = \frac{|x+y||x-y|}{4} \\ & \leq \frac{ (|x|+|y|)}{4}\cdot |x-y| \\ & \leq \frac{ 2\max \{a,b\}}{4}\cdot |x-y| \\ & = \frac{\max \{a,b\}}{2} \cdot |x-y|\end{align*}
so $f$ locally lipschitz-continuous with constant $\frac{\max \{a,b\}}{2}$.

The function $f$ is not globally lipschitz-continuous because we cannot bound from above the term $|x|+|y|$, right? (Wondering)

 

[I like Serena]

right?

Yep.

About Lipschitz-continuity:

Let $x,y\in [a,b]$ then we have that
\begin{align*}|f(x)-f(y)| & \leq \frac{ (|x|+|y|)}{4}\cdot |x-y| \\ & \leq \frac{ 2\max \{a,b\}}{4}\cdot |x-y| \\ & = \frac{\max \{a,b\}}{2} \cdot |x-y|\end{align*}
so $f$ locally lipschitz-continuous with constant $\frac{\max \{a,b\}}{2}$.

Shouldn't that be $\frac{ 2\max \{|a|,|b|\}}{4}\cdot |x-y|$? (Wondering)

The function $f$ is not globally lipschitz-continuous because we cannot bound from above the term $|x|+|y|$, right? (Wondering)

How can we be sure that is sufficient?
Shouldn't we prove that $\forall K \ge 0\ \exists x,y: \frac{|f(x)-f(y)|}{|x-y|} > K$? (Wondering)

 

[mathmari]

Shouldn't that be $\frac{ 2\max \{|a|,|b|\}}{4}\cdot |x-y|$? (Wondering)

Ah yes!

How can we be sure that is sufficient?
Shouldn't we prove that $\forall K \ge 0\ \exists x,y: \frac{|f(x)-f(y)|}{|x-y|} > K$? (Wondering)

We have the following:

\begin{align*}\frac{|f(x)-f(y)|}{|x-y|}&= \frac{|\sqrt{4+x^2}-\sqrt{4+y^2}|}{|x-y|}\\ & = \frac{|\sqrt{4+x^2}-\sqrt{4+y^2}|\cdot \frac{|\sqrt{4+x^2}+\sqrt{4+y^2}|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|}}{|x-y|} \\ & =\frac{|(4+x^2)-(4+y^2)|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|\cdot |x-y|} \\ & = \frac{|x^2-y^2|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|\cdot |x-y|}\\ & = \frac{|x+y|\cdot |x-y|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|\cdot |x-y|}\\ & = \frac{|x+y|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|}\end{align*}

How could we continue? (Wondering)

 

[I like Serena]

How could we continue?

Don't we have that $\sqrt{4+x^2}>\sqrt{x^2}=|x|$? (Wondering)

 

[mathmari]

Don't we have that $\sqrt{4+x^2}>\sqrt{x^2}=|x|$? (Wondering)

Ah! So we get the following:
\begin{align*}\frac{|f(x)-f(y)|}{|x-y|}&= \frac{|\sqrt{4+x^2}-\sqrt{4+y^2}|}{|x-y|}\\ & = \frac{|\sqrt{4+x^2}-\sqrt{4+y^2}|\cdot \frac{|\sqrt{4+x^2}+\sqrt{4+y^2}|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|}}{|x-y|} \\ & =\frac{|(4+x^2)-(4+y^2)|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|\cdot |x-y|} \\ & = \frac{|x^2-y^2|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|\cdot |x-y|}\\ & = \frac{|x+y|\cdot |x-y|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|\cdot |x-y|}\\ & = \frac{|x+y|}{|\sqrt{4+x^2}+\sqrt{4+y^2}|} \\ & < \frac{|x+y|}{|x|+|y|} \\ & <\frac{|x|+|y|}{|x|+|y|} \\ & = 1\end{align*}
Therefore we have that $|f(x)-f(y)|<|x-y|$, which means that the function is also globally lipschitz continuous, or not? (Wondering)

 

[I like Serena]

Therefore we have that $|f(x)-f(y)|<|x-y|$, which means that the function is also globally lipschitz continuous, or not?

Indeed.
We could get a clue by looking at the graph and see that the derivative is bounded to $1$ at infinity. (Nerd)

 

[mathmari]

Indeed.
We could get a clue by looking at the graph and see that the derivative is bounded to $1$ at infinity. (Nerd)

Ah ok!

Having shown that the function is globally lipschitz continuous, we don't have to show seperately that it is also locally, so we don't need the part that I showed previously, do we?

 

[I like Serena]

Having shown that the function is globally lipschitz continuous, we don't have to show seperately that it is also locally, so we don't need the part that I showed previously, do we?

Indeed. (Nod)

 

[mathmari]

Indeed. (Nod)

Great! Thank you! (Smile)