I wonder if the person asking this question might have mistaken #x rarr 1^-# (#x# approaches #1# from the left)) as "#x# approaches #-1#".
The question: "How do you evaluate the limit of #(x^2-9x)/(x^2+8x-9)# as #x rarr 1^-# (#x# approaches #1# from the left))?" is more interesting and more challenging.
#lim_(xrarr1^-)(x^2-9x)/(x^2+8x-9)=lim_(xrarr1^-)(x(x-9))/((x+9)(x-1)#
This limit has the form: #((1)(-8))/((10)(0))#, that is:
as #x rarr 1^-#, the numerator approaches #-8# and the denominator approaches #0#.
At this point we know that the limit does not exist, but we can say more.
Numbers approaching #-8# divided by numbers close to #0# are either large positive or large negative numbers. ('Large' here means 'far from #0#'.)
So, the limit does not exist because as #xrarr1^-# the values of #(x^2-9x)/(x^2+8x-9)=(x(x-9))/((x+9)(x-1)# are increasing or decreasing without bound.
As #x rarr 1^-#, the denominator approach #0# through negative values.
This is because:
for #x# a little less than #1#, the value of #x-1# is close to but a little less than #0#. (That is, the value is negative)
and #x+9# is close to #10#. So the product is negative.
Therefore, as #x rarr 1^-#, #(x^2-9x)/(x^2+8x-9)=(x(x-9))/((x+9)(x-1)# evaluates to numbers that are positive and the values are increasing as #xrarr1^-#.
Properly speaking: the limit of the expression, as #x# approaches #1# from the left, does not exist because,
as #xrarr1^-#, #(x^2-9x)/(x^2+8x-9)# increases without bound.
We express this by writing:
#lim_(xrarr1^-)(x^2-9x)/(x^2+8x-9)=oo#.
Final Note: I wanted to be thorough and correct in the details above. After you understand the principles it is cleaner and still understandable to write something like:
#lim_(xrarr1^-)(x^2-9x)/(x^2+8x-9)=lim_(xrarr1^-)(x(x-9))/((x+9)(x-1)#
#((1)(-8))/((10)(0^-))=(-8)/0^-#
So, #lim_(xrarr1^-)(x^2-9x)/(x^2+8x-9)=oo#
(Here using #0^-# to indicate a function whose values approach #0# but are #<0#)
