((x+h)2-x2)/h when h→6 is (x+h-x)(x+h+x)/h=h(2x+h)/h=2x+h (difference of two squares).
So the limit when h is near 6 is 2x+6.
I think the question should have read x→6, not h→6.
I think that we started with f(x)=x2 and we're looking for the gradient (slope) where x=6. This is the basis of differentiation.
The differential (derivative) of x2 is found by working out:
(f(x+h)-f(x))/h. Think of a right triangle on a graph. The base of the triangle is formed by two horizontal points h units apart. The height of the triangle is formed by two vertical points and the distance between them is f(x+h)-f(x) units.
So the gradient=height/base=(f(x+h)-f(x))/h. f(x)=x2 so f(x+h)=(x+h)2 and the difference is h(2x+h). When divided by h this leaves 2x+h. As h→0 (that is, h is very close to x) 2x+h gets closer to 2x. We call 2x the derivative of x2. But in this question we are looking at the gradient when x=6 so 2x=2×6=12. The gradient is very steep at x=6. Now let's rephrase the question:
The limit as h→0 when x=6 of x2, using ((x+h)2-x2)/h.