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Another limit

  1. Feb 25, 2006 #1
    Hello,I've been trying to solve [tex]\lim \frac{x+sin(x)}{2x+1}[/tex] when x tends to +∞
    I have NO clue where to start... if someone can give me any hints or help!
    We didn't learn derivatives nor l'Hopital rule yet!
     
    Last edited: Feb 25, 2006
  2. jcsd
  3. Feb 25, 2006 #2

    siddharth

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    I'll give you a hint, and let's see if you can go from there. Hint: Divide the numerator and denominator by 'x'.

    Can you see the purpose of doing this?
     
  4. Feb 25, 2006 #3

    VietDao29

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    Looking at the limit first, since you notice that:
    -1 <= sin(x) <= 1. So when x is very large (i.e it tends to infinity). sin(x), and the 1 in the denominator become negligible, so you'll be left with:
    x / (2x), right?
    But this is not rigorous. This is just what you think when you first look at the problem. It's just to have some clues to tackle the problem. If sin(x), and 1 become negligible when x tends to infinity. Why not consider to divide both numerator, and denominator by x?
    [tex]\lim_{x \rightarrow \infty} \frac{x + \sin x}{2x + 1} = \lim_{x \rightarrow \infty} \frac{\frac{x + \sin x}{x}}{\frac{2x + 1}{x}} = ...[/tex]
    Can you go from here? :)
     
  5. Feb 25, 2006 #4
    I did devide everything by 'x'
    I get [tex]\lim \frac {1+ \frac {sin(x)}{x}}{\frac {2x+1}{x}}[/tex]
    And limit of sin(x)/x is 0
    So I finally get x/(2x+1) and which is still ∞/∞
     
    Last edited: Feb 25, 2006
  6. Feb 25, 2006 #5

    HallsofIvy

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    So divide numerator and denominator of that by x!
     
  7. Feb 25, 2006 #6
    [tex]\frac {1}{\frac{2x+1}{x}}[/tex]
    ??
    Sorry... I didn't get what you mean
     
  8. Feb 25, 2006 #7

    VietDao29

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    You should note that:
    [tex]\frac{2x + 1}{x} = \frac{2x}{x} + \frac{1}{x} = 2 + \frac{1}{x}[/tex]
    [tex]\lim_{x \rightarrow \infty} \frac{x + \sin x}{2x + 1} = \lim_{x \rightarrow \infty} \frac{\frac{x + \sin x}{x}}{\frac{2x + 1}{x}} = \lim_{x \rightarrow \infty} \frac{1 + \frac{\sin x}{x}}{2 + \frac{1}{x}} = ...[/tex]
    Now can you go from here? :)
     
    Last edited: Feb 25, 2006
  9. Feb 25, 2006 #8
    Yup!!
    Thank you
     
  10. Mar 2, 2006 #9
    Hey again, i ve been tryin to solve the same one above but this time it's
    [tex]\lim \frac{x+sin(x)}{2sin(x)+1}[/tex]

    After simlifying and everything, I get to:

    lim (1+sinx/x)/(2sinx/x + 1/x)
    so if lim sin/x when x tends to +infinity, the limit above is equal to lim x, which is +infinity


    But I don't think this is the right answer, because we have to see when 2sin(x)+1 is equal to 0 and discuss each phase... don't we ? is so, how ?
     
    Last edited: Mar 2, 2006
  11. Mar 4, 2006 #10

    VietDao29

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    There is no limit there. In this limit:
    [tex]\lim_{x \rightarrow \infty} \frac{x + \sin x}{2 \sin x + 1} = \lim_{x \rightarrow \infty} \frac{1 + \frac{\sin x}{x}}{2 \frac{\sin x}{x} + \frac{1}{x}}[/tex]
    Now, as x tends to infinity, sin(x) / x tends to 0, so does 1 / x, right? So the numerator will tend to 1, whereas the denominator tends to 0.
    If the numerator tends to a constant <> 0, and the denominator tends to 0, that means the limit does not exist there.
    [/QUOTE]But I don't think this is the right answer, because we have to see when 2sin(x)+1 is equal to 0 and discuss each phase... don't we ? is so, how ?[/QUOTE]
    I don't really get this, as x tends to infinity, the numerator becomes bigger, while the numerator can only take value from -1 to 3, that means the whole expression will evaluate to some really really great number (in absolute value), ie. either ([tex]+ \infty \quad \mbox{or} \quad - \infty[/tex]) as x tends to infinity. Hence the expression does not converge to any specific number.
    Thus the limit does not exist.
    Can you get this? :)
     
  12. Mar 4, 2006 #11
    Ok makes sense!
     
    Last edited: Mar 4, 2006
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