Question

using limits...finding the tangent line of the square root of x when x=2

Answer 1

Recall that at any point (x,y) on the graph, the tangent line has slope y'

y = sqrt(x)
y' = 1/(2sqrt(x))
y'(2) = 1/(2sqrt2) = 0.35

So, now we have a point (2,sqrt2) and a slope 1/(2sqrt2)

The point-slope form of the equation gives

y-sqrt2 = 1/(2sqrt2) * (x-2)

If you want the usual slope-intercept form, rearrange stuff to get

y = 1/(2sqrt2)x + 1/sqrt2