Question

Question #2b749

Answer 1

`Lim_(xrarroo)ln(x^3+x+1)/ln(x+1)=3`

Explanation:

`Lim_(xrarroo)ln(x^3+x+1)/ln(x+1)=oo/oo`

Using L'Hopitals rule:

`Lim_(xrarroo)(1/(x^3+x+1)*(3x^2+1))/(1/(x+1))=Lim_(xrarroo)((3x^2+1)(x+1))/(x^3+x+1)`

`Lim_(xrarroo)(3x^3+3x^2+x+1)/(x^3+x+1)=Lim_(xrarroo)(x^3(3+3/x+x/x^2+1/x^3))/(x^3(1+1/x^2+1/x^3))`

`=(3+0+0+0)/(1+0+0)=3/1=3`

Answer 2

`lim_(x->oo) ln(x^3+x+1)/ln(x+1) = 3`

Explanation:

Evaluate:

`lim_(x->oo) ln(x^3+x+1)/ln(x+1)`

The limit is in the form `oo/oo` but we can use the properties of logarithms to simplify.

As the the function is defined for `x+1 > 0` we can divide and multiply the argument of the logarithm at the numerator by `(x+1)^3`:

`lim_(x->oo) ln(x^3+x+1)/ln(x+1) = lim_(x->oo) ln(((x^3+x+1)(x+1)^3)/(x+1)^3)/ln(x+1)`

Use now the properties:

`ln ( a xx b) = lna+lnb`

`ln a^n = n ln a`

to get:

`lim_(x->oo) ln(x^3+x+1)/ln(x+1) = lim_(x->oo) (ln((x^3+x+1)/(x+1)^3)+3ln(x+1))/ln(x+1)`

`lim_(x->oo) ln(x^3+x+1)/ln(x+1) = lim_(x->oo) (ln((x^3+x+1)/(x+1)^3))/ln(x+1) + 3`

Now:

`lim_(x->oo) (x^3+x+1)/(x+1)^3 = 1`

so:

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

and then:

`lim_(x->oo) ln(x^3+x+1)/ln(x+1) = 3`