Creating a tangent line from a transformation

Question

$\lim_{n\to\infty}nf(\frac xn)-nf(0)=f'(0)x$

So my intuition behind this is that the transformation of $f(x)\to nf(\frac xn)$ essentially zooms in on the graph by a factor of $n$ and this makes sense because if you think about a power series all the terms will got o zero except $a_0$ and $a_1$

But. If you apply this transformation to a general case you get any tangent line as:$$\lim_{n\to\infty}nf\big(\frac{x+(n-1)a}n\big)-(n-1)f(a)=f'(a)(x-a)+f(a)$$ which is very complicated and is difficult to get an analytic intuition to why is is specifically this.

So that's my question. From an analytic standpoint why does this work?

Here is validation but not proof for this: https://www.desmos.com/calculator/8iu4kzmgih

Answer 1

Let $h:=x/n$, then $$\lim_{n\to\infty}(nf(x/n)-nf(0))=\lim_{h\to0}\frac{f(h)-f(0)}{h}x=f'(0)x$$