Prove for Any a in R That F X is Continuous at a
Proof: Show that f(x) is continuous at a
- Thread starter shwanky
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Homework Statement
Show that f(x) is continuous at a=4
Homework Equations
[tex]f(x) = x^2 + \sqrt{7-x}[/tex]
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]
Proof 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.
Answers and Replies
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 :-/.
http://video.google.ca/videoplay?docid=8266735963828183844&q=how+to+write [Broken]
Your last line reminded me of this. :)
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 :-/.
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?
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!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]
Actually, he does have a definition for "a". Of course, he should explain what it means for a number to be "true"!
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!
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.
Are we looking for a proper epsilon/delta proof, or does your course not require proofs?
I'm not exactly sure what your asking so my answer would be, at this point no?
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?
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.
(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:
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html
http://lobe.ibme.utoronto.ca/mat196f/epsilon-delta-proofs.htm
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...
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.
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.
That was not my point. If you are using that to prove that the limit exists, you can't just assert that the two one-sided limits exist- you have to show that!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.
Are we looking for a proper epsilon/delta proof, or does your course not require proofs?
You are asked to SHOW something is true. It might not be an "epsilon-delta" proof but you are still being asked for a proof!I'm not exactly sure what your asking so my answer would be, at this point no?
How exactly would I show that they are true then?
That is up to you to decide given the requirements of your course. Look at examples done by the teacher - how do they prove things? Do they make appeals to things like 'the sum of continuous functions is continuous'?
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]
I looked in my book and the section which uses the epsilon delta proof was skipped... He said it's to complicated to for us right now and will come back to it at the end of the semester... But that really just makes this more confusing :-/...
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