# How do you evaluate #lim_(x->0^(-)) e^x/(1+lnx)#?

##### 2 Answers

Since the limit of the denominator does not readily tend to **quotient rule of limits**.

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.