What is the #lim_(x to 0^-) e^(2/x)#?

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Answer 1 · 2018-01-31T02:59:02

#lim_(x to 0^-) e^(2/x) = 0#

Given: #lim_(x to 0^-) e^(2/x)#

Push the limit through the exponential:

#e^(lim_(x to 0^-) 2/x)#

The 2 can come outside of the limit:

#e^(2lim_(x to 0^-) 1/x)#

We know that the limit goes to #-oo#:

#e^-oo = 0#

#lim_(x to 0^-) e^(2/x) = 0#