Using epsilon-delta relation prove that

[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.

[WORLD-HEN]

http://lobe.ibme.utoronto.ca/mat196f/epsilon-delta-proofs.htm

[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.

[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! :smile:

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.

[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..

[quasar987]

Oh right, thanks.

[bezgin]

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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.


Thank you very much indeed.

[tamintl]

Superb!! Helped me understand it