What is the limit as x approaches infinity of #(1+a/x)^(bx)#?

1 Answer
Aug 30, 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#)
www.schooltrainer.com