Question

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

Answer 1

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`)