I don't have df. in my notes, but is that correct?
So without looking at the solution I proceeded like this:
lim(2x-x^2)/2x
x->o
So I have factored out an x in the numerator: x(2-x)/2x
I then shortened the x, so I get
2-x/2
2-0 is 2
2/2 are then 1
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Hey Unfortunately, I don't understand b) at all. How do you even come up with these numbers? Thanks in advance for your help 🙂 Solutions:
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“Mathematic” would of course be otherwise noted (always with the “lim” in advance), but the procedure and the result would be correct.
You have a function with (re)removable definition gap at x=0. By shortening the x out, you get a straight line (temmed in normal form: f(x)=-1/2x+1), only that in this case the point (0|1) does not exist because the “un shortened” function for x=0 is not defined.