How do you find the limit of #(x^n)/(n!)# as n approaches #oo#?

1 Answer
sente
Jul 3, 2016

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

Explanation:

If #x=0# then #x^n/(n!)=0# for all #n#, meaning the limit is trivally #0#. If #x in RR# and #x!=0#, let #M# be a natural number such that #M>=2|x|#. Then

#lim_(n->oo)|x^n/(n!)| =lim_(n->oo) |x|^M/(M!)*|x|^(n-M)/(n(n-1)...(M+1))#

#=lim_(n->oo)|x|^M/(M!)*prod_(k=1)^(n-M)|x|/(M+k)#

# <= lim_(n->oo)|x|^M/(M!)*prod_(k=1)^(n-M)|x|/M#

#<=lim_(n->oo)|x|^M/(M!)*prod_(k=1)^(n-M)|x|/(2|x|)#

#=lim_(n->oo)|x|^M/(M!)*prod_(k=1)^(n-M)1/2#

#=|x|^M/(M!)*prod_(k=1)^oo1/2#

#=|x|^M/(M!)*0#

#=0#

Additionally, as #|x^n/(n!)|>0# for all #n#, we have #lim_(n->oo)|x^n/(n!)| >= 0#

By the above, we have #0 <= lim|x^n/(n!)|<=0#, implying #lim|x^n/(n!)| = 0#.

Then, as #lim_(n->oo)|a_n|=0 iff lim_(n->oo)a_n = 0#, we have #lim_(n->oo)x^n/(n!) = 0# for any #x in RR#.

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