Expert: Ahmed Salami - 11/25/2009

Question
Please help me to use the
L’Hôpital’s rule to show that lim(x→∞)x^n/e^x=0?

Answer
Hi Nara,
By L’Hôpital’s rule,
lim(x→∞)[x^n/e^x] = lim(x→∞) n.[x^(n-1)/e^x]
                 = n . lim(x→∞)[x^(n-1)/e^x]
Now,
Let L(n) = lim(x→∞)[x^n/e^x]
We can see that
L(n) = n.L(n-1)
when n = 1, we have
L(1) = 1 . lim(x→∞)[x^(1-1)/e^x]
    = 1 . lim(x→∞)[x^0/e^x]
    = lim(x→∞)[1/e^x]
    = 0
And subsequently,
L(2) = 2.L(1) = 2.0 = 0
L(3) = 3.L(2) = 3.0 = 0
and
L(n) = 0
i.e
lim(x→∞)[x^n/e^x] = 0

OR

We could have simply said, from L(n) = n.L(n-1)
L(0) = 0.L(-1) = 0
and the subsequent L's would be zero, as before.

Regards