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.
In: 8
f'(x) = lim a->0 (f(x+a) – f(x-a))/a
Being the definition of a derivative is what I assume you’re talking about. You may have seen it as
f'(x) = lim h->x (f(x)-f(h))/x-h
Both are essentially the same thing. You take a point on f(x) and you want the slope at that point. You simply do rise over run, m = Δy/Δx
That works just fine when calculating the slope between two point, but in this case we only have one.
It’s pretty easy to imagine what’s happening in the first one. We are taking the slope between two points, x+a and x-a. We just make the points infinitely close together to get the slope at x.
In the second, we do the same but we take the points x and h and we just move h infinitely close to x to get the slope at x
This is also how we get dy/dx to be the symbol for a derivative. It comes from Δy/Δx
As long as the function is continuous at x, this works fine because the limit of a function is equal to the value of the function. This is why you can’t take the derivative of a function at a noncontinuous point
Latest Answers