How can I work this limit exercise? Consider points M, N and O with coordinates (1, 0), (0, 1) and (0, 0), respectively, additionally P point with coordinates (x, y) in the graph $y = \sqrt{x}$ . Calculate:
(a) $\lim_{x \to 0^{+}} \frac{perimeter \bigtriangleup NOP}{perimeter \bigtriangleup MOP}$
(b) $\lim_{x \to 0^{+}} \frac{area \bigtriangleup NOP}{area \bigtriangleup MOP}$

One of my first ideas for this exercise, after drawing the points M, N, O and the $y = \sqrt{x}$ graph, was to see the formulas of perimeter and area for triangles.
perimeter $\bigtriangleup $: is the sum of all its three sides.
area $\bigtriangleup $: is always half the product of the height and base
However, I do not find how to use the formulas of perimeter and area of a triangle with a point P = (x, y)  in the graph $y = \sqrt{x}$.

$perimeter \bigtriangleup NOP = 1(NO) + \sqrt{x^2+(\sqrt{x}-1)^2}(NP) + \sqrt{x^2+(\sqrt{x})^2}(OP)$
$perimeter \bigtriangleup MOP = 1(MO) + \sqrt{(x-1)^2+(\sqrt{x})^2}(MP) + \sqrt{x^2+(\sqrt{x})^2}(OP)$
$area \bigtriangleup MOP = \frac{1}{2} * 1 * \sqrt{x}$
$area \bigtriangleup NOP = \frac{1}{2} * 1 * x$
I think you can finish remain.
