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Awkward limit

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data
    27. limit as x head towards infinity of xsin(1/x) is
    A 0
    B infinity
    C nonexistent
    D -1
    E 1
    from barron's ap calc 7ed, Chapter two review questions p.37
    how come the answer is E? Isn't it A b/c the limit equals infinity or neg infinity times zero? the Book explains that we can represent x sin (1/x) as sin(1/x)/(1/x). when we take the limit of this, it comes out to be E.

    29. a function is given, one with a removable disconinuity if x is not = 1 and y=4 when x = 1. The quesiton asks what is true? I. limf (x goes to 1) exhists? II. f(1) exhists. III. f is continuous at x = 1.
    The answer turns out that I is true. how come it is not I and II?

    2. Relevant equations
    limits and continuity knowledge
  2. jcsd
  3. Jan 15, 2007 #2
    First, you need to understand what a limit is. It's not the value when x *equals* zero, it's the value of the function as x *approaches* (getting arbitrarily close) to some value. Thus, you never actually have a value of zero to multiply by. For example, consider the limit as x approaches 2 of (x^2-4)/(x-2). The numerator is clearly zero if you "plug in" 2. Of course, so is the denominator. However, "plug in" a numbers such as 1.9, 1.99, 1.999, 1.9999 into the function. Clearly, the value of the function is getting closer to 4. It's easy to see this is true, because you can factor the numerator and the denominator.
    Last edited: Jan 15, 2007
  4. Jan 15, 2007 #3
    Have you learned L'Hopital's rule yet? Also, this problem is related to the limit as x approaches zero of sin(x)/x.

    Also note: the example I gave above with x^2-4 over x-2 is an example of a function with a removeable discontinuity. As you get arbitrarily close to a value of 2 for x [and this is important - on either side of 2], you'll find that the f(x) values get close to 4. In fact, you may in class take a more rigorous approach to limits, in which case, you'll determine that if I state you have to be within a billionth of the limit, can you find a close enough x-value? Since you can factor the numerator and cancel it with the denominator, this function is, in fact, *ALMOST* the same as f(x)=x+2. The difference is that in the original function, x cannot equal zero. Thus, you could rewrite the function as:
    f(x)=x+2, for all values of x except 2. At 2, you have a hole, aka removeable discontinuity. Think about where the y value is, if you're really close to 2 (either at 1.9999999995 or 2.0000000005, or even closer.)
    Last edited: Jan 15, 2007
  5. Jan 15, 2007 #4
    or use the taylor expansion of sin(1/x) and get the answer immediately.
  6. Jan 15, 2007 #5
    What's the limit of sin(x)/x as x goes to zero? Some time is spent on that in the Calculus.

    Well then, what's the difference of the limit as x goes to infinity of [tex]\frac{sin(1/x)}{1/x}[/tex]?
  7. Jan 15, 2007 #6

    Gib Z

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    Homework Helper

    I think the easiest way is re writing it is [sin(1/x)]/[1/x]. To make an easier read, let u=1/x , then as the limit x approaches infinity, u appoaches zero.

    So we figure the limit of (sin u)/u as u appraches zero. This is easy to evaluate using L'Hopitals Rule. I'm afraid I don't see how using Taylor Series will make the answer anymore obvious.

    Edit: He beat me to it..
  8. Jan 15, 2007 #7
    If I remember correctly the sin(x)/x is generally gotten from the definition of the derivative [tex]\frac{sin(x+h)-sin(x)}{h}[/tex] as h goes to zero.

    This is important since at that point we do not know what the derivative of the sin is. That may or may not bear on the matter above
  9. Jan 16, 2007 #8


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    Science Advisor

    Interesting. In those calculus books I have seen [itex]\lim_{x\rightarrow0}\frac{sin x}{x}[/itex] is derived geometrically and then used to prove the deriivative!
  10. Jan 16, 2007 #9
    HallsofIvy: Interesting. In those calculus books I have seen is derived geometrically and then used to prove the deriivative!

    I reallly meant to say the same thing. If we can handle sinx(cos(h)-1)/h, we are left with cos(x)sin(h)/h goes to cos(x), once we know the limit of sin(h)/h. (Also, we take sinx(cosx-1)/h and multiply top and bottom by cosx+1 to get [sin(x)/h][-sin^2(x)]/[(cos(h)+1)]) which arrives at 1*-0=0. Here too we employ sinx/x goes to 1.)
    Last edited: Jan 16, 2007
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