Given:
#f(x) = 1/(x + 2)#
The limit definition is
#lim_(hto0){f(x + h) - f(x)}/h#
To create #f(x + h)#, you merely substitute #x + h# for every #x#:
#f(x + h) = 1/(x + h + 2)#
Substitute these functions into the definition:
#lim_(hto0){1/(x + h + 2) - 1/(x + 2)}/h#
Multiply by 1 in the form of #{(x + h + 2)(x + 2)}/{(x + h + 2)(x + 2)}#:
#lim_(hto0){1/(x + h + 2) - 1/(x + 2)}/h{(x + h + 2)(x + 2)}/{(x + h + 2)(x + 2)}#
Multiply the numerators on the top and the #h# on the bottom:
#lim_(hto0){((x + h + 2)(x + 2))/(x + h + 2) - ((x + h + 2)(x + 2))/(x + 2)}/(h(x + h + 2)(x + 2))#
What this does is that makes the first term in the numerator become #(x + 2)# by cancelling the denominator, #(x + h + 2)#, and it makes the second term in the numerator become #x + h + 2# by canceling the denominator, #x + 2#.
#lim_(hto0){(cancel(x + h + 2)(x + 2))/cancel(x + h + 2) - ((x + h + 2)cancel(x + 2))/cancel(x + 2)}/(h(x + h + 2)(x + 2))#
Remove the canceled factors:
#lim_(hto0){(x + 2) - (x + h + 2)}/(h(x + h + 2)(x + 2))#
Distribute the minus sign in the numerator through the parenthesis:
#lim_(hto0){x + 2 -x - h - 2}/(h(x + h + 2)(x + 2))#
More canceling:
#lim_(hto0){cancel(x) cancel(+ 2) cancel(-x) - h cancel(- 2)}/(h(x + h + 2)(x + 2))#
Remove the cancelled terms:
#lim_(hto0){ - h}/(h(x + h + 2)(x + 2))#
#-h/h# becomes -1:
#lim_(hto0){ -1}/((x + h + 2)(x + 2))#
Now, it is ok to let the limit go to zero:
#{ -1}/((x + 0 + 2)(x + 2))#
Remove the 0:
#{ -1}/((x + 2)(x + 2))#
The numerator becomes a square:
#{ -1}/((x + 2)^2)#
Substitute the x coordinate for the point #(0, 1/2)# to obtain the slope, m, of the tangent line:
#m = { -1}/((0 + 2)^2)#
#m = -1/4#
