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Math Help - Limit problem

  1. #1
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    Limit problem



    I expanded everything out and got [3x^2 + x^2*sin(x)] / [x^2 + 2sin(x) + sin^2(x)] and I tried splitting the problem but it didn't work because the denominator is always 0 when I try plugging in '0' for 'x.' I can't find any way to cancel out the denominator
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  2. #2
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    Re: Limit problem

    Quote Originally Posted by PhizKid View Post


    I expanded everything out and got [3x^2 + x^2*sin(x)] / [x^2 + 2sin(x) + sin^2(x)] and I tried splitting the problem but it didn't work because the denominator is always 0 when I try plugging in '0' for 'x.' I can't find any way to cancel out the denominator
    When you expand the denominator you get

    x^2+2x\sin(x)+\sin^2(x)=x^2\left[1+2\frac{\sin(x)}{x}+\left( \frac{\sin(x)}{x}\right)^2\right]

    Now reduce.

    P.S. you should know the value of the limit

    \lim_{x \to 0}\frac{\sin(x)}{x}

    Can you finish from here?
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    Re: Limit problem

    With this kind of limit, I like to first note which factors are relevant. That (3+sin(x)) part isn't - it's going to 3, so isn't going to be the cause of the limit converging or not. The other two factors are what count - the x-squared in the numerator, and the (x+sin(x)) squared in the demoniator, are what is making this be of the form 0/0; that's where the action is; that's where you need to pull some trick or simplification to find the limit.

    As a hint, when it's "like 0/0", and has x's and sin(x)'s and is a limiti as x goes to 0, almost surely you'll end up using sin(x)/x goes to 1 as x goes to 0. You could even do the following:
    "Replace" sin(x) by (sin(x)/x)x, and then keep that (sin(x)/x) together in your algebra, as it'll be going to 1. Now you don't have anymore sin(x)'s running around - only x's and that (sin(x)/x) that's going to 1.

    For this problem, a big hint would be to observe that:

    \frac{x^2}{(x+\sin(x))^2} = \left(\frac{x}{x+\sin(x)}\right)^2 = \left(\frac{1}{1+\frac{\sin(x)}{x}}\right)^2
    Last edited by johnsomeone; September 17th 2012 at 06:20 PM.
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    Re: Limit problem

    Quote Originally Posted by johnsomeone View Post
    With this kind of limit, I like to first note which factors are relevant. That (3+sin(x)) part isn't - it's going to 3, so isn't going to be the cause of the limit converging or not. The other two factors are what count. As a hint, when it's "like 0/0", and has x's and sin(x)'s and is a limiti as x goes to 0, almost surely you'll end up using sin(x)/x goes to 1 as x goes to 0.

    As a hint, note that
    \frac{x^2}{(x+\sin(x))^2} = \left(\frac{x}{x+\sin(x)}\right)^2 = \left(\frac{1}{1+\frac{\sin(x)}{x}}\right)^2
    I don't understand how you went from \left(\frac{x}{x+\sin(x)}\right)^2 = \left(\frac{1}{1+\frac{\sin(x)}{x}}\right)^2
    This was the hint provided in the textbook, but I failed see how the numbers have suddenly all changed during this step.
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    Re: Limit problem

    \frac{x}{x+\sin(x)}

    = \left(\frac{x}{x+\sin(x)}\right) \ \frac{(1/x)}{(1/x)}

    = \frac{(x)(1/x)}{(x+\sin(x))(1/x)}

    = \frac{1}{(x)(1/x) +(\sin(x))(1/x)}

    = \frac{1}{1+\frac{\sin(x)}{x}}
    Last edited by johnsomeone; September 17th 2012 at 06:26 PM.
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    Re: Limit problem

    Quote Originally Posted by PhizKid View Post
    I don't understand how you went from \left(\frac{x}{x+\sin(x)}\right)^2 = \left(\frac{1}{1+\frac{\sin(x)}{x}}\right)^2
    This was the hint provided in the textbook, but I failed see how the numbers have suddenly all changed during this step.
    Are you saying that you don't understand how we divide numerator and denominator by x~?
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    Re: Limit problem

    Oh, I see that now. But how did you know to divide both the numerator and denominator by x?
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    Re: Limit problem

    Quote Originally Posted by PhizKid View Post
    Oh, I see that now. But how did you know to divide both the numerator and denominator by x?
    Before I wrote "As a hint, when it's "like 0/0", and has x's and sin(x)'s and is a limiti as x goes to 0, almost surely you'll end up using sin(x)/x goes to 1 as x goes to 0.".
    When I see something like this problem, after deciding that the sin(x) going to 0 is an essential reason that the limit's value is unclear, I think to myself: "I need to get an x under that sin(x)." In this case, the way to do that was to divide both numerator and demoninator by x.
    Last edited by johnsomeone; September 17th 2012 at 08:55 PM.
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