How do you evaluate #lim_(x->0^(-)) e^x/(1+lnx)#?
2 Answers
Since the limit of the denominator does not readily tend to
However, it makes more sense to evaluate the right-handed limit, since to the left of
#color(blue)(lim_(x->0^(-)) e^x/(1 + lnx))#
#=> lim_(x->0^(+)) e^x/(1 + lnx)# (where we have switched to the right-handed limit because the function is clearly continuous to the right of
#x = 0# , and at#x = 0# as well.)
#= (lim_(x->0^(+)) e^x)/(lim_(x->0^(+)) 1 + lnx)#
#= (1)/(-oo)#
#= color(blue)(0)#
This is shown below:
graph{e^x/(1 + lnx) [-1, 4, -10.78, 14.88]}
Although there is an existent limit at x = 0, it, again, doesn't make as much sense to think about how it's evaluated from the left, as it does to consider the righthand limit.
