Evaluate #lim_(x->1)((1-2x)/(x^2-1))#

Since #x=1# is a vertical asymptote (which would mean the limit as #x# approaches #1# would be #-oo#, #oo#, or does not exist), we need to check the limit from the left and right sides:

Left:

#lim_(x->1^-)((1-2x)/(x^2-1))#

When we plug in a number just slightly less than #1#, the limit would have a negative numerator and a negative denominator, therefore the limit as #x# approaches #1# from the left is #-oo#

Right:

#lim_(x->1^+)((1-2x)/(x^2-1))#

Wen we plug in a number just slightly greater than #1#, the limit would have a negative numerator and a positive denominator, therefore the limit as #x# approaches #1# from the right is #oo#

Since the limit as #x# approaches #1# from the left does not equal the limit as #x# approaches #1# from the right, we conclude that the limit does not exist.