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Function f(x) such that it's continuous

  1. Oct 14, 2009 #1
    Am I correct -

    First...define the variables -

    x – Independent variable of the function f.

    l – The “limit” f(x). The value of l is is known, arbitrary and f(x) should be equal to l.

    a – a value such that f(a) = l

    δ – The value of x can be varied by a certain amount...δ is that amount...the values it can be varied to will be δ + a to δ – a or δ – a<x< δ + a.

    ε – The value of δ has this restriction of ε (or actually it is a function of ε)...it happens that the value of l can be varied through a the value ε...or it can be l + ε to l – ε...so the value of l can be deviated by ε...δ represents this corresponding deviation by defining the value of x as such δ – a<x< δ + a.

    So if you're given a value of l and another value so as to specify how close to l do you want f(x) to be (ε), then by all this δ will have to be computed and we can say that the value of a can be varied by δ to maintain this limit of closeness to l.

    This is what I mean when I write the notion -

    and have been given separately a value of ε.

    By taking limits we assume that we're taking a section of the function f(x) such that it's continuous.

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  3. Oct 14, 2009 #2


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    Re: Limits

    One thing I would object to is "a – a value such that f(a) = l". You are assuming that the value of the function is equal to the limit- that is, that f is continuous at a which makes limits pretty much trivial.

    f(x)= x if x< 1,
    f(x)= 2 if x= 1,
    [itex]f(x)= x^2[/itex] if x> 1.

    What is [itex]\lim_{x\to 1} f(x)[/itex]?
  4. Oct 14, 2009 #3
    Re: Limits

    From what I understood about limits, it should be 2.

    However there can be 2 values of δ.................overall I'm totally confused...the 2 values of δ should be specific to the value of x, since the function behaves differently for a value less than 1 and a value more than 1.

    if value of δ has to be single for x>1 and x<1, the there should be 2 values of ε...one specifically for x>1 and the other for x<1.

    The same sort of conclusion can be drawn for any exponential, logarithmic, and I think trigonometric function also...or actually a single value of ε and δ can coexist iff the function is linear.
  5. Oct 14, 2009 #4


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    Re: Limits

    Then you need to work on your understanding of limits. Pretty much every thing you said there is wrong.

    (Don't worry too much. I regularly trip up junior and senior math majors with this. They should worry!)

    The definition of "[itex]\lim_{x\to a} f(x)= L[/itex]" is "Given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]0< |x-a|< \delta[/itex] then [itex]|f(x)- L|< \epsilon[/itex].

    If you think that the limit is 2, then I would give you an [itex]\epsilon[/itex] of, say, 0.5 and ask how close x must be to 1, that is, how small [itex]\delta[/itex] must, so that if [itex]1-\delta< x< 1+\delta[/itex] then |f(x)- 1|< 0.5. But no matter how small your [itex]\delta[/itex] is , there will exist some values of x that are strictly between [itex]1-\delta[/itex] and 1. And for those x, f(x)= x. Since x is not larger than 1, that certainly cannot be within 0.5 of 2! 2 cannot be the limit.

    I claim that 1 is the limit.

    Given any [itex]\epsilon> 0[/itex], take [itex]\delta[/itex] to be the smaller of [itex]\epsilon[/itex]/3 and 1.
    Now, suppose x is any number such that [itex]0< |x- 1|< \delta[/itex] then [itex]0< |x-1|< \epsilon/3[/itex] and 0< |x-1|< 1.

    Divide this into 2 cases:
    1)x< 1. Then x-1> 0 f(x)= x. Look at |f(x)- 1|= |x-1|. Since [itex]|x-1|< \epsilon/3[/itex] it is certainly less than [itex]\epsilon[/itex].

    2)x> 1. Then [itex]f(x)= x^2[/itex] so |f(x)- 1|= |x^2-1|= |(x-1)(x+1)|= |x-1||x+1|.
    But |x-1|< 1 so -1< x-1< 1. From that, 0< x< 2 and 1< x+ 1< 3. Since 0< 1< x+1, |x+1|= x+1< 3. [itex]|x-1||x+1|< 3|x-1|< 3\epsilon/3< \epsilon[/itex].

    That completes the proof that the limit is 1, not two.

    Notice that there is no "x= 1" case. That is because of the "0< " in the [itex]0< |x-1|< \delta[/itex] part of the definition of the limit. The value of the function at x= 1 is completely irrelevant to the limit.

    If the problem had been:
    f(x)= x if x< 1,
    f(x)= a if x= 1,
    f(x)= [itex]x^2[/itex] if x> 1.
    where a is -100 or 2000000, the limit would be exactly the same- 1.

    What you should have done is look at the "onesided limits".
    [itex]\lim_{x\to 1^-} f(x)= \lim_{x\to a} x[/itex] and there is a theorem that says "the limit, as x goes to a, of x, is a" so the limit from below is 1.
    [itex]\lim_{x\to 1^+}f(x)= \lim_{x\to 1} x^2= (\lim_{x\to 1) x)(\lim_{x\to 1)x)[/itex]
    (Another "rule of limits" says "[itex]\lim_{x\to a}f(x)g(x)= (\lim_{x\to a}f(x))(\limit_{x\to a}g(x))[/itex])
    so the limit from above is also 1. Since those two one sided limits are the same, the limit itself exists and is their common value: 1.

    That "0< |x-a|" is often neglected but is crucial because it leads to one of the most important (and also neglected) "laws of limits":
    If f(x)= g(x) for all x except a, then [itex]\lim_{x\to a}f(x)= \lim_{x\to a}g(x)[/itex]

    If you were asked to find the limit, as x goes to 2, of [itex](x^2-2)/(x-2)[/itex], you would probably do something like:
    [tex]\lim_{x\to 2}\frac{x^2- 4}{x-2}= \lim_{x\to 2}\frac{(x-2)(x+2)}{x-2}= \lim_{x\to 2} x+2= 4[/tex]
    That is completely correct, but not because the functions [itex](x^2- 4)/(x-2)[/itex] and x+2 are equal- they are not the same function. But they are equal for all x except x=2. [itex](x^2- 2)/(x-2)= (x-2)(x+2)/(x-2)[/itex] is completely correct. But (x-2)(x+2)/(x-2) is NOT. You cannot divide by 0 so that cancellation is not valid for x= 2. It is for all x except 2 so the statement about the limits is correct.

    (The graph of y= x+ 2 is a straight line. The graph of [itex]y= (x^2- 4)/(x- 2)= (x-2)(x+2)/(x-2)[/itex] is not. It is a straight line with a point missing. There is a hole at (2, 4).)
    Last edited: Oct 14, 2009
  6. Oct 15, 2009 #5
    Re: Limits

    So actually a acts as the deviant in the value of x...and the limit of deviation is called δ.

    L is the deviated value of f(x) and ε (by usual formal definition) is the limit of the amount of deviation.

    And I was wondering a as the initial value x and did not assume a variable which deviates the value of x also I thought (sort of) -

    [itex]0< |x-a|< \delta[/itex] or/and [itex]0< |x+a|< \delta[/itex]; similar for f(x), L and ε which is wrong since this will be true only for leaner functions where both sides behave the same.

    These things were wrong and I previously thought about it.

    But is it a rule that we need to take x - a only and not x + a?...considering we take only x + a -

    a and L are interrelated so f(a) = L :confused:...but this does not happen actually.

    If someone asks to find the limit of an expression as x approaches a, he's basically asking the deviated value of f(x) which will be f(a).

    I think what I've understood and stated above is wrong...a given value of a and ε at the same time will be like giving the same information twice; which is extremely unlikely.

    So I would like to confirm till now....thanks.
  7. Oct 15, 2009 #6


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    Re: Limits

    a is the given value of x at which we want to take the limit. |x-a| measures how close to a x is. For example, if a= 2 and x= 2.001, then x-a= 0.001. x+a= 2.001 tells us nothing.

    No, "a" and "[itex]\epsilon[/itex]" are completely different quantities. Assigning them values is certainly not "giving the same information twice".

    For example: If we claim that [itex]\limit_{x\to 2} 3x-1= 5[/itex] means that we can make 3x-1 as close as we please to 5 just by making x close enough to 2. A perfectly reasonable response then would be "Okay, how close to 2 does x have to be to guarentee that 3x-1 is less than .01 from 5?" In that case, a is given a 2 (the point at which we are taking the limit) and [itex]\epsilon[/itex] is given as .01 (how close we want those values to be).

    One way to answer that is to look at the distance from 3x-1 to .01 for any x. It is |3x-1-5|= |3x-6|= |3(x- 2)|= 3|x-2| and we want that to be less than .01: 3|x-2|< .01 is the same as |x- 2|< .01/3. The answer to the question is "x has to be within .01/3 of 2 in order to guarentee that 3x-1 is within .01 of 5".

    Doing exactly that with the symbol "[itex]\epsilon[/itex]" rather that the specific ".01" is exactly what we would do to prove that [itex]\lim_{x\to 2} 3x-1= 5[/itex].

    From [itex]|f(x)- a|= |3x-1-5|= |3x-6|= |3(x-2)|= 3|x-2|< \epsilon[/itex]
    so that [itex]|x-2|< \epsilon/3[/itex] and it is clear that taking [itex]\delta= \epsilon/3[/itex] or smaller is sufficient.
  8. Oct 16, 2009 #7
    Re: Limits

    Intuition can carry you quite a bit with limits once you've worked with them for a while. Except where the answer is not "obvious correct" or when it's required by a professor, no one ever thinks about epsilon-delta definitions. They are very low-level and technical.

    [tex]\lim_{x \to a} f(x) = l[/tex] simply means if I have a sequence of numbers [x1, x2, x3, ...] that get closer and closer to a, then the sequence of numbers [f(x1), f(x2), f(x3), ...] gets closer and closer to l.

    "Closer and closer" above has a precise meaning. It is actually a different kind of limit called the limit of a sequence. Limits of a sequence are more fundamental and a little easier to understand.
  9. Oct 16, 2009 #8
    Re: Limits

    It appears physics forums servers have failed today...so I will not be able to answer :(

    So a is sort of the initial value of the independent variable of f and we can call x the deviated value of a. It's wrong to assume x to be the general independent variable of x...given a question, it (x) has a specific value.

    Thus along with a, ε needs to be known to predict the value of x.

    Is this correct till now?...I'll answer the tomorrow later...today the servers are bad.
  10. Oct 17, 2009 #9
    Re: Limits

    What does |3x - 1 - 5| suppose to signify?...shouldn't it be |3x - 1 - 0.01|?

    But this does not necessarily mean f(2) = 5 or does it?...I think it has to be.

    The main problem that I'm facing here is the definition of variables...let's start with x first -

    x – General independent variable of f (from “One way to answer that is to look at the distance from 3x-1 to .01 for any x”).

    i.e it's just the name of the indpenedent variable of x...it's a socket in which various values can be plugged in (similar to when generally speaking of any function e.g g(x)) it can assume any value like a or a + δ in the same question; or in the same example x's value can change many times, for e.g -
    "In this case value of x is taken as n and here it's assumed as k."

    Finally in a question x is a variable, unlike a, l, δ and ε right?
  11. Oct 17, 2009 #10
    Re: Limits

    But what happens when we ask 'find the limit'...there has to be another value.

    I suppose limit of sequence is easier cause here this extra variable can be eliminated.
  12. Oct 17, 2009 #11


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    Re: Limits

    ???? No, the problem was to show that the limit of f(x)= 3x-1, as x goes to 2, is 5. The |f(x)- L| in the definition of limit becomes, since f(x)= 3x-1 and L= 5, |3x-1-5|. I have no idea why you think anything would be "|3x-1-0.01|"

    If f(x)= 3x- 1, then, yes, f(2)= 3(2)- 1= 5. But if you mean that any f(x) such that [itex]\lim_{x\to 2} f(x)= 5[/itex], we must have f(2)= 5, no. That is true only of functions that are continuous at x= 2. In fact, that is the definition of "continuous".

    As I said before, the fact that [itex]\lim_{x\to a}f(x)= L[/itex] is completely independent of f(a). You can change to value of f(a) to anything you like, or leave f(a) undefined. f(x) will still have the same limit, L, at a.

    It think you are wasting time on trivialities. Strictly speaking all those can be treated like variables- they can be all have differing values depending on the situation. Normally, we are "given" [itex]\epsilon[/itex] so that is normally a constant. For any given [itex]\epsilon[/itex] there exist an infinite number of possible values for [itex]\delta[/itex]. Normally a is a limit and l is given by that so those are normally constants.
  13. Oct 18, 2009 #12
    Re: Limits

    Lets just quote each of the lines and I'll type what I think about them...............cause I'm utterly confused here.
  14. Oct 18, 2009 #13


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    Re: Limits

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

    means f can be made as close to L as desired, provided x is chosen sufficiently close to a. This limit does not care about the value f(a) or any f(x) with x not near a. The ε is the maximum allowed difference between f(x) and L. The δ is a value of the difference between x and a that ensures ε will not be exceeded. In general C depends upon ε, that is δ(ε).
  15. Oct 18, 2009 #14
    Re: Limits

    the limit is 1 - first let's look at the idea behind the limit coming from the left.. since the function is defined so that for any x less than 1 is x, f(x) for any value approaching 1 is going to be x. So now imagine a line f(x) = x that suddenly stops at 1 and jumps to 2.. if you approach that point from the y=x line, of course you approach 1 and not 2. This is the intuitive idea behind it, you might ask yourself, "well, what if you approach x=1 from somewhere else? then you won't be follow the y=x line..".. well, that's what they mean by "one sided limits' - if you approach it from the left, you follow the function from the left.

    like I said before, the limit doesn't have to be f(a).. consider a certain kind of discontinuity known as a "removable discontinuity".. it can be basically a regular continuous function with just one point literally just taken out. if you take evaluate that function's limit to that point, you should get whatever you would have gotten if you did not remove that point, since the function naturally approaches that point. however, that point is not defined on that function, it has been removed.. so how could you call it f(a)? you cannot. instead, you call it L.

    "But is it a rule that we need to take x - a only and not x + a?...considering we take only x + a - "

    I think you might be confusing the notation.. x is just some point close to a. imagine you set up an interval known as delta so that any point inside that interval has a distance from a that is less than the distance between a and delta+a.
    Last edited: Oct 18, 2009
  16. Oct 19, 2009 #15
    Re: Limits

    To start off, I 'guess' the definition of the variables here are they -

    To me, [itex]0< |x-a|< \delta[/itex] proves that x like a is a constant since δ is also a constant cause it's related to ε which is a given constant.
    It's cause of ε that δ has a constant value...the value of δ is corresponding to ε.

    [itex]|f(x)- L|[/itex] will give the deviated value of f(x) to another value f(some other value)...this will also mean that x is actually a constant since if it was a variable, it would have returned infinite values for |f(x)- L|.
    Another possibility is that x can actually be a variable such that [itex]|f(x)- L|< \epsilon[/itex]; if I assume this as true, then [itex]|x-a|[/itex] will also be satisfied, that is the value of x is such that [itex]0< |x-a|< \delta[/itex].
    By definition of δ, [itex]0< |x-a|< \delta[/itex] and [itex]|f(x)- L|< \epsilon[/itex] should be corresponding...i.e if in [itex]0< |x-a|< \delta[/itex] we take a certain value of x and if this same value is applied to [itex]|f(x)- L|< \epsilon[/itex] it will satisfy this equation.

    Which one of these is true?...is x a constant or a variable which can vary it's value by a certain limit?

    Either ways, we can state that x and consequently f(x) are variables on which a limit has been posed, these values can be deviated from a and f(a) only by a certain amount of ε or δ which are a function of each other.

    I noticed that δ and ε are intervals similar to dx; so if |x-a|< δ where a is a constant, then value of x can vary from a + δ to a – δ, this has been stated here also -

    But this is causing a problem...there's just 1 value of δ and ε and the 2 values are a function of each other...what exactly is this function depends on f.

    Suppose we are considering the deviated value of the function g(y)...this deviated value is represented by suppose g(y + z) and |g(y) – g(y + z)| < ε similarly |y – z| < δ

    The values of δ and ε, in this case will be only specific to y + z cause it might (and will in almost all function) be that |g(y) – g(y + z)| != |g(y) – g(y – z)| thus |g(y) – g(y + z)| < ε and |y – z| < δ will be only specific to y + z and so -

    Will be wrong.

    Let's move on to the next set...with a broken set of dependencies.

    Then we have an example -

    f(x)= x if x< 1,
    f(x)= 2 if x= 1,
    [itex]f(x)= x^2[/itex] if x> 1.

    What is [itex]\lim_{x\to 1} f(x)[/itex]?[/QUOTE]

    Here -
    “How close x must be to 1” means that my second assertion above was true...x is a variable but it can vary value by a certain limit or within a confined interval.

    This sentence also means that a is such that f(a) = l since we are considering the function to be continuous, if a variable z is made very closer to 1, it will mean that f(z) will also be close to l and f(1) = l...this is later proved true by this -

    And we consider the function as continuous.

    This is how much the value of x can vary...this is actually the same as [tex]0< |x-a|< \delta[/tex]

    This will be right for what I stated before -

    For this, till now, whatever I've understood about limits will be right only if the function is linear.

    So basically I've been thrown into chaos...explanations through examples is crap...and asking questions like how close?...is also a bad idea for now.
  17. Oct 19, 2009 #16
    Re: Limits

    There's one major line that needs to be written before -

    "Suppose we want the value of f(x) to be confined within a certain 'limit'" of ε from L; then we can do so using this expression -

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

    And given a value of ε.

    Since the output of f(x) is controlled by x, the value of x should also has a limit such that if it's value varies within δ of 'a', it will be within L ± ε.

    Now all this doesn't mean f(a) = l...it might happen that f(a) = ∞...the relationship between a and L is that as x approaches a, f(x) approaches L.

    Am I right till this?
  18. Oct 19, 2009 #17
    Re: Limits

    In this example, what can be the limit?...as x approaches to 1, what does the value of f(x) approach to?...and answer is 1; if we state 2, it will be wrong since the function is broken...why??...cause f(x) will never exhibit values between 1 and 2 and so f(x) will never exceed 1 and be less than 2.

    Previously, I though the function was continuous...just a wrongly decoded the problem.

    Also I was assuming limits are only applicable for continuous functions.
  19. Oct 19, 2009 #18


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    Re: Limits

    If it were true that "limits are only applicable to continuous functions", then the whole subject would be trivial! For any continous function "[itex]\lim_{x\to a} f(x)= f(a)[/itex]" so the limit becomes just another name for the value of the function. Allinteresting limits are of non-continuous functions. As, for example, the most important application of limits, the derivative, [itex]\lim_{h\to 0} (f(x+h)- f(x))/h[/itex] which always has the form "0/0" and so is not continous at h= 0.
  20. Oct 19, 2009 #19
    Re: Limits

    So the definition of a function being discontinuous at a point is when the slope of that point does not exist; similarly a function is said to be continuous at a point, when it has a slope at that point.
  21. Oct 20, 2009 #20


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    Re: Limits

    That would be the definition of differentiable
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