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Math Help - limit of a function

  1. #1
    Junior Member
    Nov 2009

    limit of a function

    Our teacher is being really nice (kinda) and teaching us some principles of calculus even if she shouldn't (crappy education system in my country), so you probably understand that we don't stay much at one principle in order to learn calculus.
    The first was the limit, everything was OK until the teacher gave us an example that we didn't finish:
    limit when x->inf. of function (4*2^x + 3^x)/(3*2^x + 3^x+1).
    We always ended up to inf./inf. or 0/0, then she said (hope I translated right) we need to use base changing formulas and started to teach about slopes.
    I couldn't find anything that could be under name "base change".
    How could I solve this example? Does it have to do something with the mathematical constant e?
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  2. #2
    MHF Contributor
    Nov 2008

    You can factor out 3^x

    \frac{4 \cdot 2^x + 3^x}{3 \cdot 2^x + 3^x + 1} = \frac{3^x \left(1 + 4 \cdot \left(\frac{2}{3}\right)^x \right)}{3^x \left(1 + 3 \cdot \left(\frac{2}{3}\right)^x + \frac{1}{3^x}\right)} = \frac{1 + 4 \cdot \left(\frac{2}{3}\right)^x}{1 + 3 \cdot \left(\frac{2}{3}\right)^x + \frac{1}{3^x}}

    \left(\frac{2}{3}\right)^x and \frac{1}{3^x} have 0 for limit at +\infty

    Therefore the limit is 1
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  3. #3
    Junior Member
    Nov 2009
    Thanks for your reply, that 3^x+1 is actually 3^(x+1), I made extra space to between numbers to see that, but now I see that this forum has mathtype, I will write it down with it:
    \frac{4 \cdot 2^x + 3^x}{3 \cdot 2^x + 3^{x+1}}

    P.S. maybe the teacher gave us a wrong example, like the same thing here? She didn't seem to check her notes, just started to solve it on her own.

    wait a big, the script doesn't seem to understand this thing.
    Ok I figured out how to write equations, but I used the source of your equation and added codes together.
    It would be nice if someone told me the easy way
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  4. #4
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, regdude!

    If that final "plus one" is part of the exponent, we get a different answer.

    \lim_{x\to\infty} \,\frac{4\cdot2^x  +  3^x}{3\cdot2^x + 3^{x+1}}
    running-gag has the best approach . . .

    Factor out 3^x\!:\;\;\lim_{x\to\infty}\, \frac{3^x\left[4\left(\frac{2}{3}\right)^x + 1\right]} {3^x\left[3\left[\frac{2}{3}\right)^x + 3\right] } \;=\;\lim_{x\to\infty}\,\frac{4(\frac{2}{3})^x + 1}{3(\frac{2}{3})^x + 3}

    \text{Since }\,\lim_{x\to\infty}(\tfrac{2}{3})^x \:=\:0,\:\text{ we have: }\;\frac{4\cdot0 + 1}{3\cdot0 + 3} \;=\;\frac{1}{3}

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  5. #5
    Junior Member
    Nov 2009
    At school we tried to devide everything with 3^x * 2^x and used 3^x+1 = 3 * 3^x
    But is seems it would have given us the same result.
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