why is this not division by zero?

Question

from f.p.'s cos(x + delt x/2).sin(delt x/2) all divided by delta x/2, making a complex fraction;
as delta x approaches zero, the numerator of sin() approaches zero, but so does the denominator, (delta x/2).
Why can that apparent problem be so easily ignored?

Answer 1

this is because it is not actually 0/0. if it were at the point where this were true, then the answer would be indeterminate. however, it is asking how the graph appears as it gets closer and closer to that value, not when it is at the value itself.

Answer 2

You're taking a limit. A limit asks what does the function converge to, not what it is. If I were walking along the function from the left, I'd "expect" myself to end up at a certain point as delta-x -> 0. If I were approaching from the right, I may have some other expectation. If they're the same, then that's the limit, and we can use that expected value. The function's not being evaluated at that point, so you're not dividing by 0.

Answer 3

you can show using geometry that sin(t) / t tends to 1 as t tends to zero. Consider the unit circle in the xy plane, in the first quadrant walk to a point P which is t units along the circle ,starting at (1,0).. Join P to the origin, the angle formed has radian measure t . From P drop a line perpendicular to the x axis. You now have a triangle whose height is sin(t) and this height is certainly less than the arc of length t, sin(t)< t or sin(t) / t < 1. At (1,0) erect a perpendicular and extend the line through OP until it forms a triangle. The height of this triangle is tan(t) and that length is certainly larger than the arc t, t< tan(t) = sin(t) / cos(t) or cos(t) < sin(t) / t. As t tends to 0 then cos (t) tends to 1. So if t is close to zero we have "almost 1"

Answer 4

If you're finding the limit as it approaches zero, since both the numerator and denominator both go to zero, you can use l'Hôpital's Rule. This means that if you take the derivatives of the numerator and denominator (seperately, not as a fraction), the limit will still be the same. Since the derivative of the bottom is one half, it is not a division by zero.