0
$\begingroup$

When we talk about limits or approaching some value, we say that change dx can never be 0. it is just approaching 0. I understand that very well.

But then we say that Derivative is the slope of a tangent at the single point on graph's function.

What I don't understand is that as long as dx is some finite value (because it can't be infinitesimal small, we avoid infinity here), there will always to be two points on the graph, one approaching the another though they can be as close as we want. Now if there exists even a tiny distance between them, then how can it be that our ratio df/dx represents slope of the tangent.

Not only it represents, We say that is precisely the slope of a tangent at a single point.

In another words, Where does derivatives make there transition from approximation to the exactness ?

Question may seem silly but I can't live with this. Either Derivative is not a slope at single point or I am missing some point.

$\endgroup$
4
  • $\begingroup$ Are you comfortable with the idea of a limit becoming “infinitely close” to a value? $\endgroup$ Commented Nov 22, 2021 at 7:13
  • $\begingroup$ @Bonnaduck But the ideas of limit is to avoid that. Instead we say approaching to the value as much close we want.... $\endgroup$ Commented Nov 22, 2021 at 7:31
  • $\begingroup$ The slope of a curve at a point is the limit of the gradient of a secant line as the two points get as close as you want. $\endgroup$ Commented Nov 22, 2021 at 8:11
  • 2
    $\begingroup$ The derivative is the limit and is always exact (where defined and makes sense). The approximations which tend to the limit are always approximations and need not be exact. It is generally confusing to thing of $dx$ as a symbol which has meaning apart from in expressions such as $df/dx$ (or the integral) until the basic conceptual notions of derivative and integral are firmly grasped. $\endgroup$ Commented Nov 22, 2021 at 8:46

1 Answer 1

0
$\begingroup$

Look at a concrete example of a limit. Consider $$\lim_{x \to 1} \frac{x^2-1}{x-1}$$

This is an easy limit to figure out, since for all $x \ne 1$, $$\frac{x^2-1}{x-1} = x+1$$ Clearly the limit should be the value of $x + 1$ when $x = 1$. That is, $2$. But the fraction is not defined at $x = 1$, where the limit is being taken.

For any value of $x$ near but not equal to $1$, the value of the fraction is near, but not equal to $2$. You might put $x = 0.9$ or $0.99$ or $0.999$ or ... into the fraction to get values of $1.9$ or $1.99$, or $1.999$ or .... These are approximations to the limit. But by its definition the limit itself is not an approximation. Instead, it is that value that is being approximated. Since the closer the input $x$ is to $1$, the closer the output value is to $2$. The limit is not any of the approximations. It is instead $2$ itself, a number you cannot get out of the fraction, since $x$ cannot be $1$.

Limits do not "approach some value". The function is what approaches the value. The limit is the value being approached. Taking a limit can be thought of as a process. But the limit itself is not the process, but the result you get from the process.

This concept does not need to involve infinities or infinitesimals. The standard definition of $$\lim_{x \to a} f(x)$$ is that it is the unique number $L$ which satisfies the condition

For every $\epsilon > 0$, there is some $\delta > 0$ such that for all $x$ with $0<|x - a| < \delta$, we have $|f(x) - L| < \epsilon$.

No mention of infinity or infinitesimals or processes in that definition. The limit is just some number $L$ that meets this condition. If there is no such number, we say the limit does not converge. If there is such an $L$, then that is the limit (you can prove from the condition that there can only be one such number).

Derivatives are limits. If $dx \equiv x-a$ is finite, no matter how small, then in general you do not get the derivative. All you get is an approximation. The derivative is the value those approximations are approximating.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.