I only ever encountered the limit while learning derivation by first principles in calculus. I understood all the theory behind first principles, but we were never told what happens to the limit h -> 0. Our teacher just said that it goes away after we divide by h, and that’s all I got.
I understand that the limit h -> 0 represents the gap between x and (x + h) getting smaller and smaller. But how does this gap disappear at the end? From searching online I’ve learned that limits are not *equality*, h never *equals* zero, it just gets closer and closer to it. But then why does it equal zero at the end? How is h -> 0 no longer intrinsic to f'(x)? This might be a dumb question but it has stumped me for ages now.
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The formal way of thinking about limits is called the epsilon-delta construction. The idea is that, even if we can’t define the behaviour exactly at the point, we can look at its behaviour as we get closer and closer. More specifically, when we say that the limit as x approaches some point c of some function f(x) approaches some value L (or lim x->c f(x) = L), what we mean is that for any small positive value (which we call ε, or epsilon) we can find another small positive value (which we call δ, or delta) such that for all the x-values closer than δ to c (i.e. 0 < |x – c| < δ), f(x) is closer than ε to L (i.e. |f(x) – L| < ε). We say that 0 < |x – c| because if it were *equal* to 0, we’d be at the point c, and that might not give a well-defined value for f(x).
The way you’d actually do this in practice is to pick δ to be some appropriate function of ε and then find f(x) for 0 < |x – c| < δ, and see whether those are within ε of L. For example, let’s take the derivative of x^(2). We know that this is lim h->0 ((x+h)^2 – x^(2))/h and we can expand brackets and rearrange to get lim h->0 (2xh + h^(2))/h.
Now, we can’t just put in h = 0, because that would get us 0/0, which is undefined, and we also can’t cancel the h out and then set it to 0, because when you cancel out a variable you intrinsically assume it’s not 0, and if we cancelled h out and then set h to 0 we’d be contradicting ourselves. Let’s use the ε-δ approach. Now, we know that the limit here should be 2x, so that’s our value of L, and we’re approaching 0, so that’s our value of c, so we want to find some small positive value δ such that for 0 < |h| < δ, |f(h) – 2x| < ε for all small positive values ε.
First, let’s look at what f(h) is. We have f(h) = (2xh + h^(2))/h. Now, and this is the clever bit, because we said that |h| is greater than 0, that means h is not 0, which means it’s totally ok to cancel it out. That gives us f(h) = 2x + h, so |f(h) – 2x| = |2x + h – 2x| = |h|. If |h| < δ, then we have |f(h) – 2x| < δ, and we want this to be less than ε, so we just have to choose something like δ = ε/2. ε/2 is always smaller than ε for positive ε, so we’ve successfully found a δ-value that guarantees we’ll be within ε of 2x. Therefore, the limit is 2x, and at no point did we do anything illegal like dividing by 0 or getting an undefined expression like 0/0.
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