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

1 Answer
Alan P.
Oct 22, 2016

#lim_(xrarr2^+)(x^2-8)/(8x-16) = color(green)(-oo)#

Explanation:

Given:
#color(white)("XXX")lim_(xrarr2^+)(x^2-8)/(8x-16) #

If #xrarr2^+# then #x > 2#
and therefore
#color(white)("XXX")(8x-16) > 0#
(i.e. #8x-16# is positive as #xrarr2^+#)

At #x=2# (and thus as #xrarr2^+#)
#color(white)("XXX")x^2-8=-4#

Therefore
#color(white)("XXX")lim_(xrarr2^+) (x^2-8)/(8x-16)#

#color(white)("XXX")="non-zero, negative value"/("positive value approaching " 0)=-oo#

We could also see this by graphing the given expression:
enter image source here

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