What is # lim_(->-oo) f(x) = absx/(16-x^2) #?

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Answer 1 · 2015-11-21T17:23:58

#lim_(x rarr -oo)|x|/(16-x^2) = 0#

Since we're dealing with #-oo# we can say #|x| = -x#, so

#lim_(x rarr -oo)|x|/(16-x^2) = lim_(x rarr -oo)-x/(16-x^2)#

Now, with that out of the way, factor out an #x# on the denominator
#lim_(x rarr -oo)-x/(16-x^2) = lim_(x rarr -oo)x/(x(16/x - x))#

Cancel the two #x#s otu

#lim_(x rarr -oo) = 1/(16/x - x)#

Now, because that #16/x# will become a #0# in the limit, which will make the limit as a whole become a #0#, or

#lim_(x rarr -oo) 1/(0 - x) = lim_(x rarr -oo) 1/x = 0#