Limits in Mathematics?

245 views

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

12 Answers

Anonymous 0 Comments

> 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?

It doesn’t. Let’s work an example: let’s compute the derivative of f(x) = x^2 from first principles.

—–

By definition, f'(x) = lim(h->0) f(x-h)-f(x) / h

Since our function is f(x) = x^(2), we can write this out explicitly: lim(h->0) (x-h)^2 – x^2 / h, where I’m writing without parens with the understanding that everything left of the slash is the numerator.

Expand out (x-h)^2 to get x^2 – 2xh + h^(2). Then the numerator is x^2 – 2xh + h^2 – x^(2), and the x^(2)’s cancel. So our numerator is 2xh + h^(2), all over h.

Since we are considering values near, but not equal to, zero, h != 0. And that means it’s valid to cancel the common factor of *h* on both top and bottom. That gets us 2x + h.

So, what’s lim(h->0) 2x + h? Well, this is a continuous function in *h*, so **now** we are allowed to plug in h = 0 (because by definition, the limit of a continuous function at a point is just the value of the function there). We couldn’t do this before because it would have resulted in 0/0, an indeterminate form that doesn’t give us anything useful.

h is never actually 0 here. But the continuity of the function we got at the end makes the value of the limit equal to what the value *would be* if h *were* 0 (and if the original expression didn’t have a removal discontinuity at 0).

You are viewing 1 out of 12 answers, click here to view all answers.