How do you find the limit of #(x^2 - 8) / (8x-16)# as x approaches #2^-#?

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Answer 1 · 2016-07-01T10:46:33

#= +oo#

#lim_{x to 2^-} (x^2 - 8) / (8x-16)#

if we let #x = 2 - epsilon#, where #0 < epsilon < < 1#

then we have

#lim_{epsilon to 0} ((2 - epsilon)^2 - 8) / (8(2 - epsilon)-16)#

#lim_{epsilon to 0} (4 - 4 epsilon + epsilon^2 - 8) / (-8 epsilon)#

#lim_{epsilon to 0} (-4 - 4 epsilon + epsilon^2 ) / (-8 epsilon)#

#lim_{epsilon to 0} 1/(2 epsilon) + 1/2 - epsilon/8 #

#= +oo# as #0 < epsilon < < 1#