We have that: $$\begin{align} \lim_{x \to 2} x^2 &= 4\\ D_{x = 2}\left(x^2\right) &= 4 \end{align}$$
In both cases the result is $4$. So limit and derivative are always the same?
If not, what is the relation (if any exists)?
3 Answers
$\begingroup$$\lim_{x \to 3} x^2 = 9$ and $\left(\frac{d}{dx} x^2\right)_{x=3} = 6$, which are not the same. When you mention "limit", you are really talking about the value of the function $f(x) = x^2$; in general, this is not the same as the value of its derivative.
However, derivatives and limits are certainly related. The derivative of a function $f(x)$ at $x_0$ is defined by $$f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h},$$ when this limit exists.
$\endgroup$ $\begingroup$The fact that in your example the limit and the derivative are equal is only a matter of coincidence.
You can understand the relation between a limit and a derivative if you look at their definition. First, the concept of limit is defined: $$\lim_{x \to \alpha} f(x) = l \quad\iff\quad \forall\, U(l),\ \exists\,V(\alpha)\quad\text{s.t.}\quad x \neq \alpha \in V(\alpha) \implies f(x) \in U(l)$$ where the notation $U(x_0)$ denotes the neighborhood of $x_0$. Also note that in your example the limit to $x_0$ is just $f(x_0)$, but this only holds for continuous functions. In general, we are not interested in the function's behaviour at the point we are evaluating the limit. This is reflected in the definition, as the $x \neq \alpha$ shows.
After you define what a limit represents you can define derivatives: $$f'(x_0) = \lim_{h \to 0}\frac{f(x_0 + h) - f(x_0)}h \in \mathbb R.$$
As you can see a derivative is defined on top of the limit concept, so this is the relation between the two.
$\endgroup$ $\begingroup$Limit of $x^2$ at $x=2$ and Derivative of $x^2$ at $x=2$ in both case result is $4$
Correct.
so is it always same limit and derivative ?
No, take any other number $x = a$, the limit is going to be $a^2$, the derivative is $2a$, and these aren't equal unless $a = 2$.
In general, the only function that satisfies $f'(x) = f(x)$ for all $x$ is $f(x) = Ce^x$.
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