How do you find the limit of #((x^2)-x-2)/(x-1)# as x approaches #1^+#?

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Answer 1 · 2016-12-12T20:28:18

See below.

#lim_(xrarr1^+)(x^2-x-2) = -2#

#lim_(xrarr1^+)(x-1) = 0^+#

So,

#lim_(xrarr1^+)(x^2-x-2)/(x-1)# has form #(-2)/0^+#.

As #xrarr1^+#, the quotient is decreasing without bound.

We write:

#lim_(xrarr1^+)(x^2-x-2)/(x-1) = -oo#