Why is the limit as #x->2# for #3/(2-x)# undefined?

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Answer 1 · 2016-10-31T06:11:22

You can use the properties of logarithms to check. Starting with #lim_(x->2) e^(3/(2-x))#:

#ln(lim_(x->2) e^(3/(2-x))) = lim_(x->2) ln(e^(3/(2-x)))#

#= lim_(x->2) 3/(2-x)#

But this function is of the form #1/(x)#, so let us examine the limit from both sides...

#=> lim_(x->2^+) 3/(2-x) = -oo#
#=> lim_(x->2^-) 3/(2-x) = oo#

You can see that here:

graph{3/(2-x) [-5.46, 14.54, -5.12, 4.88]}

Thus, undoing the #"ln"#:

#e^(lim_(x->2^+) 3/(2-x)) = e^(-oo) = 0#
#e^(lim_(x->2^-) 3/(2-x)) = e^(oo) = oo#

i.e. the limit from both sides at once is undefined.