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Using epsilon-delta relation prove that

  1. Nov 25, 2004 #1
    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. jcsd
  3. Nov 25, 2004 #2
  4. Nov 25, 2004 #3

    quasar987

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    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.
     
  5. Nov 26, 2004 #4

    quasar987

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    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 [itex]a[/itex] is L if f(x) approaches L more and more as x approaches [itex]a[/itex] more and more.

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

    In mathematical language, we write: consider [itex]f:\mathcal{D} \rightarrow \mathbb{R}[/itex] a function and [itex]a[/itex] an accumulation point of its domain [itex]\mathcal{D}[/itex]. We say that f as L for a limit at point [itex]a[/itex] if for any given positive real number [itex]\epsilon[/itex], there exists a positive real number [itex]\delta[/itex] such that for all x element of [itex]\mathcal{D}[/itex] and such that [itex]0<|x-a|<\delta[/itex], f(x) is such that [itex]|f(x)-L|<\epsilon[/itex] (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

    [tex]\lim_{x \rightarrow a} f(x) = L[/tex]

    The [itex]0<|x-a|<\delta[/itex] and [itex]|f(x)-L|<\epsilon[/itex] parts only means " [...] such that for all x element of the domain and in the interval [itex]]a-\delta,a+\delta[ \setminus \{a\}[/itex], f is indeed confined in the interval [itex]]L-\epsilon, L+\epsilon[[/itex]. This is because for all real numbers y, z with z > 0 , [itex]|y| < z \Leftrightarrow -z < y < z \Leftrightarrow y \in ]-z,z[[/itex]. Substitute y by [itex]x-a[/itex] (respectively [itex]f(x)-L[/itex]) and z by [itex]\delta[/itex] (resp. [itex]\epsilon[/itex]) and you got your inequality (while taking care of preserving the [itex]0<|x-a|[/itex] 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: [itex]f:\mathcal{D} \rightarrow \mathbb{R}[/itex] a function and [itex]a \in \mathcal{D}'[/itex]. We say that f as L for a limit at point [itex]a[/itex] if [itex]\forall \epsilon>0, \ \exists \delta>0[/itex] such that [itex]x \in \mathcal{D} \cap V'(a,\delta) \Rightarrow f(x) \in V(L,\epsilon)[/itex].


    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 [itex]f(x) = \sqrt{16-x}[/itex], [itex]\mathcal{D} = ]-\infty,16][/itex] (because negative roots are not defined in [itex]\mathbb{R}[/itex]), a is 7, and L is 3. So, following our definition, we wish to see if for all numbers [itex]\epsilon>0[/itex], we can find a number [itex]\delta>0[/itex] such that [itex]x \in \mathcal{D}[/itex] and [itex]0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon[/itex]. 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 [itex]\delta[/itex] FOR ALL [itex]\epsilon[/itex]? We simply say "consider an arbitrary number [itex]\epsilon>0[/itex]". If we can show that there exists a [itex]\delta[/itex] for this epsilon, it will be true for all of them, since our epsilon is not specified!

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

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

    [tex]|\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}|[/tex]

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

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

    Now what [itex]\delta[/itex] makes it so that [itex]0<|x-7|<\delta \Rightarrow |x-7|< \epsilon[/itex]? I believe [itex]\delta = \epsilon[/itex] does the work! :smile:

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

    [tex]\lim_{x \rightarrow 7} \sqrt{16-x} = 3[/tex]

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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 [itex]\epsilon[/itex] and [itex]\delta[/itex]. Good luck.
     
    Last edited: Nov 26, 2004
  6. Nov 26, 2004 #5

    arildno

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    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
  7. Nov 26, 2004 #6

    quasar987

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    Oh right, thanks.
     
  8. Nov 26, 2004 #7

    Thank you very much indeed.
     
  9. Dec 13, 2011 #8
    Superb!! Helped me understand it
     
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