Why does a derivative exist on a linear function?

Question

I understand that for a linear function, like $f(x)=3x$, the derivative at point $x$ would be $3$, but I don't understand why. The derivative's slope is equal to the tangent line's slope, but I was thinking that the tangent line shouldn't exist on a linear function since it would have to touch the graph at more than $1$ point in order to be parallel at a certain point. At point $(0,0)$ on the aforementioned $f(x)$, the derivative would have a constant line of $f'(x)=3$, but that's not perpendicular to the point.

Answer 1

The definition of the derivative of a function $f$ at a point $x$ is given by $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$$ In this case, we have $$f'(x) = \lim_{h \to 0} \frac{3(x + h) - 3x}{h} = \lim_{h \to 0} \frac{3h}{h} = \lim_{h \to 0} 3 = 3.$$ The limit exists, so the function is differentiable. Note how there's no implicit assumptions about how the linearisation intersects the function. There is no requirement that the tangent line intersects a function exactly once.