By definition, the slope of the tangent line to the graph of a given function #f(x)# at a given point #x_0# is the derivative #f'(x_0)#, which in turn is defined as follows:
#f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}#
In your case, #f(x) = sqrt(x+2)# and #x_0 = 2#. The definition becomes
#f'(2) = \lim_{h \to 0} \frac{sqrt((2+h)+2)-sqrt(2+2)}{h} = \lim_{h \to 0} \frac{sqrt(4+h)-2}{h}#
To solve this limit we may "razionalize" the expression:
#\frac{sqrt(4+h)-2}{h} = \frac{sqrt(4+h)-2}{h} \cdot \frac{sqrt(4+h)+2}{sqrt(4+h)+2} = \frac{(4+h)-4}{sqrt(4+h)+2} = \frac{h}{sqrt(4+h)+2}#
The limit becomes
#f'(2) = \lim_{h \to 0}\frac{h}{h(sqrt(4+h)+2)} = \lim_{h \to 0}\frac{1}{sqrt(4+h)+2} = \frac{1}{sqrt(4+0)+2} = \frac{1}{2+2} = \frac{1}{4}#
