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

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Answer 1 · 2016-10-01T22:08:25

#-oo#

#lim_(x to 1^-) 1/(x-1)+e^(x^2)#

Well, #lim_(x to 1^-) e^(x^2) = e# so we can lift that term out

#= e + lim_(x to 1^-) 1/(x-1)#

if we then define #h = x-1# where #0 < abs h " << " 1#

The limit becomes
#e + lim_(h to 0^-) 1/h#

#= e + lim_(h to 0) - 1/abs h# because h is negative as we are approaching the limit from the left

and # lim_(h to 0) 1/abs h = oo#

#implies - oo#