#lim_{x to -4} (1/4x + 1/x) / (4+x)# and clearly that is #(-5/4)/0# at x = -4
we can let #x= -4 + h#, so #h = x+4#, where #0<|h| < < 1#. This allows us to explore the effect of small variations #h# about #x = -4#
so the limit becomes
#lim_{h to 0} (1/4(-4+h) + 1/(-4+h)) / (4-4 + h)#
#lim_{h to 0} (-1 + h/4 + 1/(-4+h)) / (h)#
we can expand this part of the numerator ie
# 1/(-4+h) = (-1/4)/(1 - 1/4h) = - 1/4 (1 - h/4)^{-1}#
The binomial expansion is
#-1/4 (1 - (-1) h/4 + mathcal{O} (h^2)) #
#-1/4 - h/16 + mathcal{O} (h^2)) #
so the limit is now
#lim_{h to 0} (-1 + h/4 -1/4 - h/16 + mathcal{O} (h^2)) / (h)#
#lim_{h to 0} ( color{red}{-5/4} - (3h)/16 + mathcal{O} (h^2)) / (color{red}{h}) #
IOW:
if h < 0, ie we are to the left of x = -4, the limit is #oo#
if h > 0, ie we are to right of x = -4, the limit is #-oo#
