How do you find #lim (1-x)/(2-x)# as #x->2#?

1 Answer
Dec 16, 2016

The limit does not exist.

Explanation:

As #xrarr2#, the numerator goes to #-1# and the denominator goes to #0#. The limits does not exist.

We can say more about why the limit does not exist.

As #x# approaches #2# from the right (through values greater than #2#), #2-x# is a negative number close to #0# (a negative fraction).
So, as #xrarr2^+#, the quotient #(1-x)/(2-x)# increases without bound.
We write #lim_(xrarr2^+)(1-x)/(2-x) = oo#

As #x# approaches #2# from the left, #2-x# is a positive number close to #0#.
So, as #xrarr2^-#, the quotient #(1-x)/(2-x)# decreases without bound.
We write #lim_(xrarr2^-)(1-x)/(2-x) = -oo#