Question

Evaluate the limit `lim_(x->0) (e^x+3x)^(1/x)`?

Answer 1

`e^4`

Explanation:

As we know `e^x = 1+x+O(x^2)` then

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

now making `y = 4x`

`lim_(x->0) (e^x+3x)^(1/x) =lim_(y->0)(1+y)^(4/y) = (lim_(y->0)(1+y)^(1/y))^4 = e^4`

Answer 2

`lim_(x->0) (e^x+3x)^(1/x) = e^4`

Explanation:

Write the function as:

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

Consider now the limit:

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

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

`lim_(x->0) ln(e^x+3x)/x = lim_(x->0) (d/dx ln(e^x+3x))/(d/dx x)`

`lim_(x->0) ln(e^x+3x)/x = lim_(x->0) (e^x+3)/(e^x+3x) = 4`

As the limit is finite and the function `e^x` is continuous for `x in RR` we have:

`lim_(x->0) e^(ln(e^x+3x)/x) = e^((lim_(x->0) ln(e^x+3x)/x)) = e^4`