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Proof: Show that f(x) is continuous at a

  1. Mar 6, 2007 #1
    I'm a little nervous about a test I have on Thursday and I was wondering if this is adequate for a proof of the following equation.

    1. The problem statement, all variables and given/known data
    Show that f(x) is continuous at a=4

    2. Relevant equations
    [tex]f(x) = x^2 + \sqrt{7-x}[/tex]

    3. The attempt at a solution

    For f(x) to be continuous at a = 4
    a) f(a) must be defined

    b) [tex]\lim_{x \to a} f(x)[/tex] must exist

    c) [tex]\lim_{x \to a} f(x) = f(a)[/tex]


    1. [tex]f(a) = f(4) = 4^2 + \sqrt{7-4} = 16 + \sqrt{3}[/tex]
    Therefore, f(a) is defined at a=4.

    2. [tex]\lim_{x \to a} f(x) = \lim_{x \to 4} x^2 + \sqrt{7-x}[/tex]

    [tex]\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

    [tex]\lim_{x \to 4^-} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

    Since [tex]\lim_{x \to 4^+} x^2 + \sqrt{7-x} = 16 + \sqrt{3} = \lim_{x \to 4^-} x^2 + \sqrt{7-x} = \lim_{x \to 4} f(x)[/tex] exists.

    [tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex]

    3. By steps 1 and 2 we know that f(4) = [tex] 16 + \sqrt{3}[/tex] and [tex]\lim_{x \to 4} x^2 + \sqrt{7-x} = 16 + \sqrt{3}[/tex].

    Since [tex]\lim_{x \to 4} f(x) = 16 + \sqrt{3} = f(4)[/tex], [tex]f(x) = x^2 + \sqrt{7-x}[/tex] is continuous at a = 4.
    Last edited: Mar 6, 2007
  2. jcsd
  3. Mar 6, 2007 #2
    adequate? more than enough,in my times if a curve went to the same number from the -ve side and the +ve side, then it would be continous:)
  4. Mar 6, 2007 #3

    matt grime

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    That is not adequate. State your definition of continuity and show it is true. I saw nothing there that constitutes a proof. Nothing there even acknowledges the definition of continuous so it cannot possibly be a proof.
  5. Mar 6, 2007 #4
    Do I have to write the definition each time I prove the continuity of an equation, or rather should I?
  6. Mar 6, 2007 #5
    According to my book, a function f is said to be continuous at [tex]\lim_{x \to a} f(x) = f(a)[/tex]. Following this definition it says that that a function is continuous at a point if a, b and c are true :-/.
    Last edited: Mar 6, 2007
  7. Mar 6, 2007 #6
    http://video.google.ca/videoplay?docid=8266735963828183844&q=how+to+write [Broken]

    Your last line reminded me of this. :)
    Last edited by a moderator: May 2, 2017
  8. Mar 6, 2007 #7
    sad part is, at one point I was considering being an english major.... MUYAY! o_O
  9. Mar 7, 2007 #8

    matt grime

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    What are a,b and c? What does it mean for f(x) to tend to f(a)? Are we looking for a proper epsilon/delta proof, or does your course not require proofs?
  10. Mar 7, 2007 #9


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    But how do you get those last two limits? Not by just setting x= 4, surely, since that requires knowing that the function is continuous at 4- exactly what you are trying to prove!
  11. Mar 7, 2007 #10


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    Actually, he does have a definition for "a". Of course, he should explain what it means for a number to be "true"!
  12. Mar 7, 2007 #11
    There's a theorem in my book which states a [tex]\lim_{x \to a} = L[/tex] exists if and only if [tex]\lim_{x \to a^+} f(x) = L = \lim_{x \to a^-} f(x)[/tex] unfortunately, my book doesn't substatiate this with a proof. It just states the theorem and my professor didn't bother proving this.

    I'm not exactly sure what your asking so my answer would be, at this point no?
  13. Mar 7, 2007 #12
  14. Mar 7, 2007 #13

    Just take it to be a defintion of a limit, the limit exists if and only if the two one-sided limits exist and are equal.

    Let f be a function defined on an open interval containing c (except posibly at c) and let L be a real number. The statement

    [tex]\lim_{x \to c} f(x) = L[/tex]

    means that for each [tex]\epsilon > 0[/tex] there exists a [tex]\delta > 0[/tex] such that if

    [tex]0 < |x-c| < \delta[/tex], then [tex]|f(x) - L| < \epsilon[/tex]

    which basically says if for any small positive number [tex]\epsilon[/tex] with which you say "Can we make f(x) lie in the interval (L-[tex]\varepsilon[/tex] , L+[tex]\varepsilon[/tex])" by finding a number [tex]\delta[/tex] such that we make x lie in the interval (c-[tex]\delta[/tex], c+[tex]\delta[/tex]) for all positive [tex]\varepsilon[/tex]If the answer is yes then the limit is L as x approaches c and if f(c)=L then the function is continuous at c.

    But thats the formal definition of the limit, and might not be used in high schools for example. But I think that is how you ussually prove continuity, with an epsilon-delta proof... But if you haven't seen this stuff before, maybe phone a friend ( smart one preferably :P ) and ask if you are supposed to use the regular limit laws to 'prove' the limits exist.
    Last edited: Mar 7, 2007
  15. Mar 7, 2007 #14
    If you don't mind me asking, what textbook are you using? Epsilon and delta represent small positive numbers in this scenario. (but they are just variables, right?)

    (this part was in reference to a question posed which has since been deleted..)

    Try looking at some of these if you need help with epsilon-delta proofs:


    And if you have questions about it feel free to ask.

    (edit: and make sure its neccesary for your course. Although the precise definition of a limit gave me a better idea of what a limit really is, and it probably won't hurt to learn it. If you don't need it with your course, just show through the regular limit laws (the ones you were using in your first post) that the left and right hand limits exist and are equal to L, and that f(c)=L) In your proof maybe suggest why you are finding the limits (to prove that both one-sided limits exist and are equal) and such...
    Last edited by a moderator: Apr 22, 2017
  16. Mar 7, 2007 #15
    Is this a freshman-level course or a upper-level undergraduate course? If it is a freshman-level course, then your proof is probably adequate, but if it is an upper-level undergraduate course, then you probably should use epsilon-delta to prove the limit.
  17. Mar 7, 2007 #16
    freshman... for the love of God freshman ^.^... hehe... I hope at some point this thing starts making some sense...
  18. Mar 7, 2007 #17
    Yea, keep in mind that I was just introducing you to the idea of a delta-epsilon proof because you seemed interested in what it was.

    edit: then you were on the right path (if you don't need the delta-epsilon 'proof') :) Just maybe mention in your proof what you are doing and why before you do it.
    Last edited: Mar 7, 2007
  19. Mar 7, 2007 #18
    I'm currently using Stewart "Calculus" 5th ed, Brooks/Cole. I dont' know if we get to epsilon delta proofs further along in the book, but for now they aren't discussed in any of the chapters that have been assigned.
  20. Mar 7, 2007 #19
    I think its in the appendix. Its optional.

    Actually I have a copy and mine introduces it right away, but I know there are other copies where its in the appendix. :P

    If you don't need it, don't do it with delta-epsilon proofs. I don't know what you need to do, only you do, or at least should. :P If your teacher hasn't even mentioned delta-epsilon proofs, I'd say your safe to just apply your limit laws. Anyways, im out, good luck. Cya.
    Last edited: Mar 7, 2007
  21. Mar 7, 2007 #20
    I can't believe I have 3 years of this... -.-, I think I'm gonna die. lol
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