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

Answer 1

Emma Aldridge

#-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#

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Answer 2

Scarlett Allison

To evaluate the limit of 1/(x-1) + e^(x^2) as x approaches 1^-, we can substitute the value of x into the expression. However, this would result in division by zero in the first term, which is undefined. Therefore, we cannot directly evaluate the limit using substitution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.