Question

Question #a4e1a

Answer 1

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`.