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Math Help - limit with g/L

  1. #1
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    limit with g/L

    A tank contains 3000 L of pure water. Brine that contains 25 g of salt per liter of water is pumped into the tank at a rate of 25 L/min. The concentration of salt after t minutes (in grams per liter) is given by the function below. As t , what does the concentration approach?



    you have to find g/L
    so i thought you plug in 25 for t, but when i did that i just got 3000, which is wrong
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by kyleu03 View Post
    A tank contains 3000 L of pure water. Brine that contains 25 g of salt per liter of water is pumped into the tank at a rate of 25 L/min. The concentration of salt after t minutes (in grams per liter) is given by the function below. As t , what does the concentration approach?



    you have to find g/L
    so i thought you plug in 25 for t, but when i did that i just got 3000, which is wrong
    I may be reading this wrong, and I probably am because I'm really tired, but...

    It seems to me that the only relevant info here is the given function. To find C(t) as t\to\infty simply divide numerator and denominator by t. Then the limit becomes obvious.

    \lim_{t\to\infty}\frac{25t}{120+t}=\lim_{t\to\inft  y}\frac{25}{\frac{120}{t}+1}=25g/L
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