Question #644fe

1 Answer
sente
Dec 1, 2015

Apply the squeeze theorem to show that
#lim_(x->oo)1/xln(x^x)/(x!) = 0#

Explanation:

First, we'll do some algebraic manipulation inside the limit:

#lim_(x->oo)1/xln(x^x)/(x!) = lim_(x->oo)1/x(xln(x))/(x!)#

#= lim_(x->oo)ln(x)/(x!)#

Now there are various approaches to evaluating this limit. Intuition immediately tells us it should be #0# as the factorial function grows much faster than the log function, however we can make a stronger argument using the squeeze theorem.

The squeeze theorem states that
if #g(x) <= f(x) <=h(x)#
and
#lim_(x->a)g(x) = lim_(x->a)h(x) = L#
then
#lim_(x->a)f(x) = L#

The way to think about it is that if you can bound a function above and below and its bounds converge to the same value, then the function must also converge to that value.

First, let's make the upper bound by noting that #ln(x) < x# for #x > 0#, as
#e^x > x#
#=> ln(e^x) > ln(x)#
#=> x > ln(x)#

Therefore
#ln(x)/(x!) < x/(x!) = 1/((x-1)!)#

For our lower bound, note that for #x > 0#
#ln(x) > 0# and #x! > 0#
#=> ln(x)/(x!) > 0#

So we have #0 < ln(x)/(x!) < 1/((x-1)!)#

But #lim_(x->oo)1/((x-1)!) = lim_(x->oo)0 = 0#

Thus, by the squeeze theorem,

#lim_(x->oo)ln(x)/(x!) = 0#

As this is the same as our original function, we get the result
#lim_(x->oo)1/xln(x^x)/(x!) = 0#

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