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Algebraically Solving a Limit

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Algebraically solve the following limit, show all work:
    $$\lim_{x\to0^-} \frac{e^x\cos(x)}{x}$$
    2. Relevant equations
    I don't know exactly how to go about doing this, but I think I can use the Squeeze Theorem somewhere in here:
    ##g(x) \leq f(x) \leq h(x)##
    If ##\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L## then ##\lim_{x\to a} f(x) = L##

    3. The attempt at a solution
    I started off using the laws of limits to separate the limit into a product of limits and simplify it:
    ##\lim_{x\to0^-} \frac{e^x\cos(x)}{x} = \lim_{x\to0^-} e^x \times \lim_{x\to0^-} \frac{\cos(x)}{x} = 1 \times \lim_{x\to0^-}\frac{\cos(x)}{x}##
    Then, I started looking at using the squeeze theorem to solve this:
    ##-1 \leq \cos(x) \leq 1 \longrightarrow \frac{-1}{x} \leq \frac{\cos(x)}{x} \leq \frac{1}{x}##
    This doesn't work, since one limit approaches minus infinity and the other approaches plus infinity.

    This is where I'm at. Any tips as to what I can do next? Or do I need to forget the squeeze theorem and try to solve this differently? I'm at least confident in my work up to eliminating ##e## from the equation.
     
    Last edited: Oct 13, 2016
  2. jcsd
  3. Oct 13, 2016 #2

    Mark44

    Staff: Mentor

    The above isn't valid. After you take the limit, x shouldn't appear any more.
    Try the squeezer theorem on the whole thing -- ##\frac{e^x \cos(x)}x##. Near x = 0, ex is close to 1.
     
  4. Oct 13, 2016 #3
    Whoops, totally meant to leave the limit in for ##\frac{\cos(x)}{x}##, and just evaluate ##e^x## to 1. Fixed that in my OP. That's something I can do, right?

    When you say to use squeeze theorem on the entire thing, do you mean for me to look for functions ##g(x)## and ##h(x)## s.t. ##g(x) \leq \frac{e^x\cos(x)}{x} \leq h(x)## or to start with ##-1 \leq \cos(x) \leq 1## and transform that to ##\frac{-e^x}{x} \leq \frac{e^x\cos(x)}{x} \leq \frac{e^x}{x}##? In the latter situation I hit the same hurdle (limits go to plus and minus infinity), and for the former, I can't think of appropriate functions ##g(x)## and ##h(x)##.
     
  5. Oct 13, 2016 #4

    Mark44

    Staff: Mentor

    On second thought, I don't think my suggestion is helpful. Have you learned about Maclaurin series yet? If so, you can write the first few terms of the Maclaurin series of ##e^x\cos(x)## by multiplying a few terms of the series for ##e^x## by a few terms of the series for ##\cos(x)##. Then look at the quotient of that series divided by x.
     
  6. Oct 13, 2016 #5
    Unfortunately I haven't learned anything about sequences and series yet, though I do need to start learning it. I do know that this was meant to be solved with no complex knowledge of sequences and series or higher level calculus, though (asked in Calc I class, and we've just gone over limits and derivatives thus far).
     
  7. Oct 13, 2016 #6

    Mark44

    Staff: Mentor

    Ok, looking at the entire expression, ##\frac{e^x\cos(x)}{x}##, what happens to the numerator as ##x \to 0^-##? What happens to the denominator as ##x \to 0^-##?
     
  8. Oct 13, 2016 #7
    I'm a bit confused at this point how to incorporate that I'm approaching zero from the negative side. I see graphically why it makes a difference, but here, it doesn't seem that important?

    As ##x\to0^-##, ##e^x## approaches 1, and ##cos(x)## approaches 1. So, the numerator approaches 1, right? The denominator, obviously, approaches zero as ##x\to0^-##, since it is simply ##x##. How can I use this to come to a solution?
     
  9. Oct 13, 2016 #8

    Mark44

    Staff: Mentor

    Calculate values for ##\frac{e^x\cos(x)}x## using x values getting successively closer to 0 through the negative numbers, such as -.1, -.01, -.001, and so on. Does this suggest what ##\frac{e^x\cos(x)}x## is doing?

    This should give you a good idea of what the limit is, but I don't know what they have in mind for you to "algebraically find the limit," the problem requirement stated in post #1.
     
  10. Oct 13, 2016 #9
    If I'm recalling correctly, the problem statement specifically told me not to use tables or graphs (sorry, I should've been more accurate in my OP - I don't have the problem statement at the moment, so I was paraphrasing from what I recalled).

    I graphed this in my calculator, and it appears that the limit approaches minus infinity. I'm playing around with different intervals in the table on my calculator, and in general it seems that as I approach zero from the negative side it does, in fact, approach minus infinity.

    Any other ideas? I'll play around with this a bit longer, but I might just update this thread next week when I get a chance to talk with my professor.
     
  11. Oct 13, 2016 #10

    Mark44

    Staff: Mentor

    Negative infinity is correct.
     
  12. Oct 14, 2016 #11

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What happens if x tends to infinity through positive x values?
    When does a function have limit at a certain x value ? Has this function limit at x=0?
     
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