Question

What is `Lim_(x-> 0) (1+3x)^(1/(2x))` by using L'hopital/L'hospital rule?

`Lim_(x-> 0) (1+3x)^(1/(2x))`

Answer 1

`lim_(x->0)(1+3x)^(1/(2x)) = e^(3/2)`

Explanation:

Write the function as:

`(1+3x)^(1/(2x)) = (e^(ln(1+3x)))^(1/(2x)) = e^(ln(1+3x)/(2x))`

and consider the limit:

`lim_(x->0) ln(1+3x)/(2x)`

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

`lim_(x->0) ln(1+3x)/(2x) = lim_(x->0) (d/dx ln(1+3x))/(d/dx (2x)) = lim_(x->0) (3/(1+3x))/2= 3/2`

As the exponential function is continuous for any `x in RR`:

`lim_(x->0)e^(ln(1+3x)/(2x)) = e^(lim_(x->0) ln(1+3x)/(2x)) = e^(3/2)`