How do you find the limit of #(x-3) /( x-4)# as x approaches #4^-#?

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Answer 1 · 2017-11-23T00:10:24

#lim_(x->4^-) (x-3)/(x-4) = -oo#

When #3 < x < 4# we have:

#0 < (x - 3) < 1#

#-1 < (x - 4) < 0#

So:

#(x-3)/(x-4) < 0#

since it is the quotient of a positive numerator and negative denominator.

When #x=4#, we have:

#(x-3)/(x-4) = 1/0#

Since this has a non-zero numerator and zero denominator, the function #(x-3)/(x-4)# has a vertical asymptote at #x=4# and we find:

#lim_(x->4^-) (x-3)/(x-4) = -oo#

graph{(x-3)/(x-4) [-7.88, 12.12, -4.16, 5.84]}