f(x) = 3 - 3/3-2^(negative infinity)

how do you solve this algebraically

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- Feb 17th 2011, 01:54 PMFallenStar117Calculate:
f(x) = 3 - 3/3-2^(negative infinity)

how do you solve this algebraically - Feb 17th 2011, 02:01 PMtopsquark
Solve what? Do you have to simplify this? Is it some kind of limit? Numerically if what you have is

$\displaystyle \displaystyle \frac{3 - 3}{3 - 2^{-\infty}}$

the value is 0. (But even that expression has problems because $\displaystyle \displaystyle 2^{-\infty}$ does not have an actual value.)

-Dan - Feb 17th 2011, 02:52 PMsinx
You will get 0 over some number that gets infinitely smaller in the denominator, which will simplify to 0.

It's not really a solution though. This seems, like topsquark suggested, to be some description of a limit problem.