# Using epsilon-delta relation prove that

1. Nov 25, 2004

### bezgin

I have a serious problem with understanding the definition of limits.

Prove that Lim(x->7) Sqrt(16-x)=3

I'd be grateful if you could explain why you do each step when you solve this question. Thanks.

2. Nov 25, 2004

### WORLD-HEN

Last edited by a moderator: Apr 21, 2017
3. Nov 25, 2004

### quasar987

I would love to do it but don't have the time now.

I will do it tomorrow if no one has done it since.

4. Nov 26, 2004

### quasar987

I remember when I was trying to understand this stuff, I could use any information I could get so I also included an explanation on what the definition means additionally to the line of reasoning to follow when solving the problem.

Intuitively, we would say that the limit of a function f(x) at a point $a$ is L if f(x) approaches L more and more as x approaches $a$ more and more.

This is what the official definition is trying to say. By chosing an arbitrary number $\epsilon$, we define a certain range $]L-\epsilon, L+\epsilon[$ of numbers around L and we ask: is there a range $]a-\delta,a+\delta[ \setminus \{a\}$ such that for all x in that range, f(x) is in $]L-\epsilon, L+\epsilon[$? This definition meets our intuitive idea of a limit when we think of $\epsilon$ as being as small as we can imagine. It becomes: is it true that there is a range around $a$ such that for all x in that range, f(x) is as near to L as we want?

In mathematical language, we write: consider $f:\mathcal{D} \rightarrow \mathbb{R}$ a function and $a$ an accumulation point of its domain $\mathcal{D}$. We say that f as L for a limit at point $a$ if for any given positive real number $\epsilon$, there exists a positive real number $\delta$ such that for all x element of $\mathcal{D}$ and such that $0<|x-a|<\delta$, f(x) is such that $|f(x)-L|<\epsilon$ (it is important to regard f(x) as the image of x by f, i.e. the number associated to x, not as the function/transformation/rule f in general) and we write

$$\lim_{x \rightarrow a} f(x) = L$$

The $0<|x-a|<\delta$ and $|f(x)-L|<\epsilon$ parts only means " [...] such that for all x element of the domain and in the interval $]a-\delta,a+\delta[ \setminus \{a\}$, f is indeed confined in the interval $]L-\epsilon, L+\epsilon[$. This is because for all real numbers y, z with z > 0 , $|y| < z \Leftrightarrow -z < y < z \Leftrightarrow y \in ]-z,z[$. Substitute y by $x-a$ (respectively $f(x)-L$) and z by $\delta$ (resp. $\epsilon$) and you got your inequality (while taking care of preserving the $0<|x-a|$ condition). (This might seem evident to you but I remember I couldn't see it back then)

Finally, the definition can also be written in ultra compact form as: $f:\mathcal{D} \rightarrow \mathbb{R}$ a function and $a \in \mathcal{D}'$. We say that f as L for a limit at point $a$ if $\forall \epsilon>0, \ \exists \delta>0$ such that $x \in \mathcal{D} \cap V'(a,\delta) \Rightarrow f(x) \in V(L,\epsilon)$.

N.B. Notice that the definition of limit does not require that the point a itself be an element of the domain. That is to say, the definition does not require that f(x) approaches f(a) as x approaches a! f(a) may or may not be defined. Actually, if L = f(a) we say that the function is continuous at the point a of its domain.

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Now, for that particular problem: we see that $f(x) = \sqrt{16-x}$, $\mathcal{D} = ]-\infty,16]$ (because negative roots are not defined in $\mathbb{R}$), a is 7, and L is 3. So, following our definition, we wish to see if for all numbers $\epsilon>0$, we can find a number $\delta>0$ such that $x \in \mathcal{D}$ and $0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$. This is the form of the definition that is most practical to work with when solving those kind of problems.

So how could we prove that there exists a such $\delta$ FOR ALL $\epsilon$? We simply say "consider an arbitrary number $\epsilon>0$". If we can show that there exists a $\delta$ for this epsilon, it will be true for all of them, since our epsilon is not specified!

So, step one is to write: "Consider $\epsilon>0$. We wish to find a number $\delta > 0$ such that $x \in ]-\infty,16]$ and $0<|x-7|<\delta \Rightarrow |\sqrt{16-x} - 3|<\epsilon$." How we're gonna do that? We're going to try to find a relation between $\epsilon$ and $\delta$ that makes this implication true.

First step in this particular problem is to multiply $\sqrt{16-x} - 3$ by its conjugate:

$$|\sqrt{16-x} - 3| = |\sqrt{16-x} - 3 \frac{\sqrt{16-x} + 3}{\sqrt{16-x} + 3}| = |\frac{7-x}{\sqrt{16-x} + 3}| = |\frac{x-7}{\sqrt{16-x} + 3}|$$

(because for any real number y, |y| = |-y|)

Therefore, finding a $\delta$ such that $0<|x-7|<\delta \Rightarrow |\sqrt{16-x} - 3|<\epsilon$ is the same as finding a $\delta$ such that $0<|x-7|<\delta \Rightarrow |x-7/\sqrt{16-x} + 3|<\epsilon$. And now we have almost won. We first have to realise that for all x of the domain, $\sqrt{16-x} + 3 \geq 3$, which implicates that $|x-7 / \sqrt{16-x} + 3| < |x-7|$. Now since $|x-7 / \sqrt{16-x} + 3| < |x-7|$, if we can find a $\delta$ such that $0<|x-7|<\delta \Rightarrow |x-7|< \epsilon$, it will also be true for this same $\delta$ that $0<|x-7|<\delta \Rightarrow |x-7 / \sqrt{16-x} + 3|<\epsilon$ (because for all real numbers w, y, z, w<y and y<z ==> w<z).

Now what $\delta$ makes it so that $0<|x-7|<\delta \Rightarrow |x-7|< \epsilon$? I believe $\delta = \epsilon$ does the work!

Therefor for any given $\epsilon>0$, we have a corresponding $\delta>0$ that meets the requirements set by the definition so that we can write

$$\lim_{x \rightarrow 7} \sqrt{16-x} = 3$$

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In most limit problems, the strategy is to use algebraic inequalities on |f(x) - L| so it takes the form |x - a|, so we can set a relation between $\epsilon$ and $\delta$. Good luck.

Last edited: Nov 26, 2004
5. Nov 26, 2004

### arildno

quasar: You ought to edit your definition so that L may exist even if f(a) does not equal L!
That is, the limit-checking process is concerned about the behaviour in a PUNCTUATED neighbourhood of "a", not at "a" itself..

Last edited: Nov 26, 2004
6. Nov 26, 2004

### quasar987

Oh right, thanks.

7. Nov 26, 2004

### bezgin

Thank you very much indeed.

8. Dec 13, 2011

### tamintl

Superb!! Helped me understand it