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What is the limit of #x/e^x# as x approaches infinity?

1 Answer
Jul 19, 2016


#lim_(x->∞) x/e^x = 0#


Since direct substitution yields an indeterminate form of #∞/∞#, we can conclude that we can apply L'Hospital's rule, which basically involves taking a derivative of the numerator and the denominator.

In our case, we have

#lim_(x->∞) x/e^x = lim_(x->∞) 1/e^x = 0#

This makes sense because if we look at the graph of the original function, we can see that the function clearly approaches #0# as #x->∞#.

graph{x/e^x [-1.856, 5.072, -1.588, 1.877]}