Question

Determine the limit: limit of (1+3h)^1/h as h approaches 0. Help!?

Answer 1

`lim_(h->0) (1+3h)^(1/h) = e^3`

Explanation:

Given the function `f(h) = (1+3h)^(1/h)` consider the function:

`g(h) = ln(f(h)) = ln ( (1+3h)^(1/h)) = 1/hln(1+3h)`

so that: `f(h) = e^(g(h))`

The limit:

`lim_(h->0) ln(1+3h)/h`

is in the form `0/0` so we can apply l'Hospital's rule:

`lim_(h->0) ln(1+3h)/h = lim_(h->0) (d/(dh) ln(1+3h))/(d/(dh) h)`

`lim_(h->0) ln(1+3h)/h = lim_(h->0) 3/(1+3h) = 3`

As `e^x` is a continuous function for every `x in RR`:

`lim_(h->0) f(h) = lim_(h->0) e^(g(h)) = e^(lim_(h->0) g(h)) = e^3`