Question #a4e1a

1 Answer
Aug 31, 2017

As #xrarr-oo#, the exponent of #e^-x# approaches #+oo#, since the minus signs will cancel one another out.

In the denominator, values will approach #-oo# but at a much slower rate than which #e^-x# will approach #+oo#.

The exponential function heavily outweighs just #x#, so the function will approach some infinity and not level off at #0#.

Note, however, that #e^-x>0# for all values of #x#, and as #xrarr-oo#, we see that #x<0#. Since the values in the denominator are negative, the overall function will approach #-oo#.

Thus, #lim_(xrarr-oo)e^-x/x=-oo#.