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Homework Help: Some concrete limits

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Hello! I got few questions, about limits.
    º [tex]\lim_{x \rightarrow 0}(\frac{sin(x)}{x})=1[/tex]
    If I take values for x close to zero I get:
    f(x)=sinx/x
    f(0.1)=0.017453283
    f(0.01)=0.017453292
    as I can see it is not even close to 1.
    What is the problem? Where I am doing wrong?

    º [tex]\lim_{x \rightarrow \infty}(\frac{1}{x})=0[/tex]
    Now, for all integers I agree that [tex]\lim_{x \rightarrow \infty}(\frac{1}{x})=0[/tex] (thanks to HallsofIvy for [itex]\infty[/itex]), but what for 1/2, 1/3, 1/4 ?
    [tex]\lim_{x \rightarrow \infty}(\frac{1}{x})=\frac{1}{1/4}=4[/tex] and not 0 ?

    º [tex]\lim_{x \rightarrow 0^-}(e^{\frac{1}{x}})[/tex]
    How will I find the bound of the expression above?

    2. Relevant equations


    3. The attempt at a solution
    [tex]\lim_{x \rightarrow 0^-}(e^{\frac{1}{x}})[/tex]
    I understand that x<0 (so the values for x are tending to 0 from the left side), and
    [tex]\lim_{n \rightarrow \infty}(x_n)=0[/tex]

    For example, I know how to find the bound for:
    [tex]\lim_{x \rightarrow 2^+}(\frac{x}{x-2})[/tex]
    D=R\{2}
    xn>2
    [tex]\lim_{n \rightarrow \infty}(x_n)=2[/tex]
    [tex]x_n-2>0[/tex]
    [tex]\lim_{n \rightarrow \infty}(x_n-2)=0[/tex]
    so that:
    [tex]\lim_{x \rightarrow 2^+}(\frac{x}{x-2})=\lim_{n \rightarrow \infty}(\frac{x_n}{x_n-2})=\frac{2}{\lim_{n \rightarrow \infty}(2-x_n)}=+\infty[/tex]

    Thanks in advance.
     
  2. jcsd
  3. Nov 20, 2008 #2

    Mark44

    Staff: Mentor

    Are you using a calculator to do these? If so, I think your calculator is in degree mode. It needs to be in radian mode. As the values of x get smaller, the value of your expression will get closer to 1.
    The limit is for x growing very large, so you shouldn't concern yourself with small values of x. On the other hand,
    [tex]\lim_{x \rightarrow 0^+}(\frac{1}{x})=\infty[/tex]
    which is more related to what you're doing with 1/2, 1/3, and so on.
    As x approaches 0 from the negative side, 1/x approaches neg. infinity, so e^(1/x) approaches 0. Do you need more explanation than that?
     
  4. Nov 21, 2008 #3
    Thanks for the post Mark44.

    Yes, I was using calculator in degree mode. Now with radian mode everything is all right.

    For the second one. Sorry, I wasn't so clear. I was learning about the number "e". So for one task (example):
    [tex]\lim_{x \rightarrow \infty}(1+\frac{1}{x-3})=\lim_{x \rightarrow \infty}(1+\frac{1}{x-3})^{(x-3)+3}=\lim_{t \rightarrow \infty}(1+\frac{1}{t})^t*\lim_{t \rightarrow \infty}(1+\frac{1}{t})^3=e*(1+0)^3=e[/tex]
    As we can see they put [tex]\lim_{t \rightarrow \infty}(\frac{1}{t})=0[/tex]. How is this possible? What about for t=1/2,1/3,1/4 ? It wouldn't be zero in that case.

    Mark44, sorry for misunderstanding again. Yes I understand all of that, but how will I "show" or "prove" that. Aren't there any calculations?
    Thanks in advance.
     
  5. Nov 21, 2008 #4

    Mark44

    Staff: Mentor

    For your first limit above, I think you are missing an exponent of x. In other words, I think it should be:
    [tex]\lim_{x \rightarrow \infty}(1+\frac{1}{x-3})^x[/tex]
    Now I think I understand what you're saying. The first limit was as x approached infinity and involved an expression with (x - 3). They substituted t = x - 3 and changed the limit variable from x to t (as x gets very large, so does t).

    In the third limit expression (before they took the limit), there are two factors:
    [tex]}(1+\frac{1}{t})^t[/tex] and
    [tex]}(1+\frac{1}{t})^3[/tex].
    The second one is straightforward to evaluate in the limit, and turns out to be just 1. If you multiply it out before taking the limit, you have 1 + 3*1/t + 3*1/t^2 + 1/t^3, which approaches 1 as t gets large.

    The first one is more tricky, and you can't just say that 1/t approaches 0 as t gets large. There are two competing effects going on: the base, 1 + 1/t, is getting closer to 1, but the exponent t is getting larger. The net effect is that (1 + 1/t)^t approaches the number e as t gets large. If I recall correctly, one of the definitions of e is precisely this limit.

     
  6. Nov 21, 2008 #5
    Now I understand. I misjudged [itex]\infty[/itex].

    But for [tex]\lim_{x \rightarrow 0^-}(e^{1/x})[/tex], I need to "show", "explain" the result of the limit. How will I do that?

    Regards.
     
  7. Nov 21, 2008 #6

    HallsofIvy

    User Avatar
    Science Advisor

    For x very close to 0, but negative, say -0.00001, 1/x is -100000, a very large negative number. What is e-100000? what is e any very large negative number?
     
  8. Nov 21, 2008 #7
    Thanks for the post. It definitely will tend to zero.

    And what about [tex]\lim_{x \rightarrow 0^+}(\frac{1}{1+e^{1/x}})[/tex]?

    [tex]\lim_{x \rightarrow 0^+}(e^{1/x})=\infty[/tex] ?
     
  9. Nov 21, 2008 #8

    Mark44

    Staff: Mentor

    For the first one, as x approaches 0 (from the right), 1/x grows without bound (approaches infinity), so 1 + 1/x also grows without bound, which makes the fraction approach 0.

    For the second, that's the right value.
     
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