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# Evaluate lim as x approaches negative infinity, 1/e^x?

Rewrite and use limits at infinity for ##e^u##.

##lim_(urarroo)e^u=oo## ##" "## and ##" "## ##lim_(urarr-oo)e^u=0##

##1/e^x = e^-x##

As ##xrarr-oo##, the exponent ##-x rarr oo##,so we have the first case listed above.

##lim_(xrarr-oo)1/e^x = oo##