How do you use L'hospital's rule to find the limit #lim_(x->oo)(1+a/x)^(bx)# ?

1 Answer
Sep 11, 2014

By using logarithm and l'Hopital's Rule,
#lim_{x to infty}(1+a/x)^{bx}=e^{ab}#.

By using the substitution #t=a/x# or equivalently #x=a/t#,
#(1+a/x)^{bx}=(1+t)^{{ab}/t}#

By using logarithmic properties,
#=e^{ln[(1+t)^{{ab}/t}]}=e^{{ab}/t ln(1+t)}=e^{ab{ln(1+t)}/t}#

By l'Hopital's Rule,
#lim_{t to 0}{ln(1+t)}/{t}=lim_{t to 0}{1/{1+t}}/{1}=1#

Hence,
#lim_{x to infty}(1+a/x)^{bx}=e^{ab lim_{t to 0}{ln(1+t)}/{t}}=e^{ab}#

(Note: #t to 0# as #x to infty#)