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Limit Problems

  1. Dec 8, 2014 #1
    (I have posted this in this section, rather than homework, because I hope to improve my general understanding of methods of finding limits through these problems.)

    1: [tex]\lim_{x\rightarrow0} (\frac{cosec(x)}{x^3} - \frac{sinh(x)}{x^5})[/tex]

    I don't really know what to do with this one. I tried differentiating a few times but it doesn't seem to work. Rewrite cosec(x) as sin(x) and then Taylor series, maybe, but I can't get it to work usefully - which terms to neglect and why?

    2: [tex]\lim_{x\rightarrow0} (\int_{x}^{\frac{\pi}{2}} (\frac{ycos(y)-sin(y)}{y^2}) dy)[/tex]

    The solution equates this expression to

    [tex]\lim_{x\rightarrow0} (\int_{x}^{\frac{\pi}{2}} (\frac{d}{dy} (\frac{sin(y)}{y})) dy)[/tex]

    from which the value of the limit follows easily. But what are the steps to calling these expressions equal?

    3: My textbook also notes that the limit is distributive over addition, it shows that lim(f(x)+g(x)) = lim(f(x)) + lim(g(x)) and lim(f(x)*g(x)) = lim(f(x)) * lim(g(x)). It then says that this does not hold if you end up with a 0/0 or infinity/infinity or infinity*0. But even with these exceptions the rules still seem sketchy - can't you end up with infinity-infinity, which is not a valid result? How should I modify the rules to show when it safe to split the limits like this, and when it is not?
  2. jcsd
  3. Dec 8, 2014 #2

    Stephen Tashi

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    Confusion is understandable. Rules in most calculus books are not written with mathematical precision. They are written as handy guides to working typical exercises.

    A definition of the limit [itex] lim_{x\rightarrow a} f(x) = L [/itex] is introduced. Later in the book, other definitions are introduced that also use the terminology "limit" and the notation "lim" such as [itex] \lim_{x\rightarrow \infty} f(x) = L [/itex] and [itex] lim_{x \rightarrow a} f(x) = \infty [/itex] and [itex] lim_{x \rightarrow \infty} f(x) = \infty [/itex] and [itex] lim_{x \rightarrow a} f(x) = -\infty [/itex] etc. The upshot of this is that term "limit" is ambiguous. If you see a statement about "limits", you always have to consider what kind of limits the text is talking about.

    For example if there is a question about whether [itex] lim_{x \rightarrow a} f(x) [/itex] exists, you must consider than if [itex] lim_{x \rightarrow a } f(x) = \infty [/itex] then the answer may be "The limit does not exist" because the question might only refer to limits of the type [itex] \lim _{x \rightarrow a} f(x) = L [/itex] for a finite [itex] L [/itex].

    When reading a rule like [itex] lim_{x\rightarrow a} (f(x) + g(x)) = lim_{x\rightarrow a} f(x) + \lim_{x\rightarrow a} g(x) [/itex] you have to pay attention to the words that accompany it. (You can't do mathematics just by working with symbols.) A conservative textbook author might use the words "when all the limits involved exist", meaning when all the limits are of the form [itex] lim_{x \rightarrow a} h(x) = L [/itex] with [itex] a [/itex] and [itex] L [/itex] finite. A more daring author might use words that include the cases of other types of limits.

    I don't know anyone who has memorized a set of rules broad enough to deal with simple operations on all possible combinations limits, as the term "limit" is variously defined. In spite of the way some instructors enjoin students to think rigorously, the way people deal with combinations in practice is to use intuition. For example if [itex] lim_{x \rightarrow a} f(x)= 2 [/itex] and [itex] lim_{x\rightarrow a} g(x) = \infty [/itex] people use intuition to conclude [itex] \lim_{x \rightarrow a} (f(x) + g(x)) = \infty [/itex]. If [itex] lim_{x \rightarrow a} f(x)= -\infty [/itex] and [itex] lim_{x\rightarrow a} g(x) = \infty [/itex] people use intuition to conclude they can't predict whats going on with [itex] \lim_{x \rightarrow a} (f(x) + g(x)) [/itex] without further analysis.
  4. Dec 8, 2014 #3

    Stephen Tashi

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    I'm not sure what you're asking.

    Are you asking the general question:
    "If [itex] lim_{x \rightarrow a} \int_x^b f(y)dy = L [/itex] and [itex] g(y) = f(y) [/itex] then what rule tells me
    [itex] lim_{x \rightarrow a} \int_x^b g(y)dy = L ?" [/itex] ?
  5. Dec 8, 2014 #4


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    For this one, write it in the more simple ##\frac{1}{(\sin x) x^3}-\frac{e^{x}-e^{-x}}{2x^5}##
    Plugging in ##x\to 0## should not leave much doubt about where this goes.
    This step hinges on recognizing the fraction as a derivative of a fraction (quotient rule). ##\frac{d}{dy}\frac{\sin y}{y} = \frac{y \cos y - sin y }{y^2}. ##
    You could end up with infinity minus infinity...if that happens, you should rearrange some terms and see if you can simplify it that way.
    L'hopital's rule works well for fractions, but if you are adding and subtracting limits, sometimes you need to use inequalities to show that one is much bigger than the other. If they scale the same way to infinity, you can sometimes recombine the terms and come up with something that looks more reasonable. In general, the rules for splitting limits are based on the idea that the limit exists. When you are not sure if it exists, you can a) assume it does and see what happens, or b) manipulate the expressions first.
  6. Dec 8, 2014 #5
    Thanks for the help on the other parts.

    But I don't see how this rearrangement takes us to the solution - we end up with 1/0-0/0 i.e. infinity minus an undefined number. Don't know where that will go ...

    (And the answer actually isn't anything as simple as 0 or infinity anyway ...)

    What do you mean by "words that include the cases of other types of limits"?

    Is the rule as restricted as you say, that it requires both [itex] a [/itex] and [itex] L [/itex] finite? Would it not hold in all cases where [itex] L [/itex] is finite for both limits? What if L is 0?

    I understand the point about intuition but I would like to consider at least a little bit more detail before giving up.
  7. Dec 8, 2014 #6


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    For 1) what if you were to write it as ## \frac{ \frac{x^2}{\sin x}+e^{-x}-e^x}{x^5}## Then take 5 derivatives of the top and bottom by L'Hopital's rule to see what is left?
    I imagine that you will end up with something like ##-\frac{2}{5!}##.
  8. Dec 10, 2014 #7

    Stephen Tashi

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    I mean words like "where [itex] a [/itex] is an extended real number" that would let the rule apply to limits involving [itex] x \rightarrow \infty [/itex] or [itex] x \rightarrow -\infty [/itex]. Or words like "where [itex] L_1 [/itex] and [itex] L_2 [/itex] are both finite or both infinite in the same sense".

    Which rule are we talking about? The property "Is finite" applies to a number that is zero. Zero is a finite number.
  9. Dec 10, 2014 #8


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    Expanding each term as a Taylor series works well here.
    $$\frac{\csc x}{x^3} = \frac{1}{x^3\sin x} = \frac{1}{x^4} \cdot \frac{1}{1-\left(\frac{x^2}{3!}-\frac{x^4}{5!}+\cdots\right)} = \cdots$$
  10. Dec 10, 2014 #9

    Ray Vickson

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    For 1, expand both terms in series before taking any limits. We have:
    [tex] \rm{cosec}(x) = \frac{1}{x} + \frac{1}{6}x +\frac{7}{360} x^3+\frac{31}{15120}x^5+\frac{127}{604800}x^7 + O(x^9) [/tex]
    [tex] \sinh(x) = x + \frac{1}{6}x^3 + \frac{1}{120}x^5 + \frac{1}{5040}x^7 + O(x^9) [/tex]
    so, for small ##x \neq 0## we have
    [tex] \frac{\rm{cosec}(x)}{x^3} - \frac{\sinh(x)}{x^5} = \frac{1}{90} + \frac{1}{540}x^2 + O(x^4) [/tex]
    The limit ##x \to 0## is easy to get from this.

    To get the first series, write
    [tex] \rm{cosec}(x) = \frac{1}{\sin(x)} = \frac{1}{x(1-y)}, \;\; y = \frac{1}{6}x^2 -\frac{1}{120}x^4 + \frac{1}{5040} x^6 + \cdots[/tex]
    and then expand ##1/(1-y) =1+ y + y^2 + y^3 + \cdots##
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