Q. `lim_(x->oo)` `[(2)^(x+1) + (3)^(x+1)] / [(2)^x + (3)^x]` = ?

### 1 Answer | Add Yours

You need to force factor `3^(x+1)` to numerator and `3^x` to denominator, such that:

`lim_(x->oo) (3^(x+1)(2^(x+1)/3^(x + 1) + 1))/(3^x(2^x/3^x + 1))`

Since `a^x/b^x = (a/b)^x` yields:

`lim_(x->oo) (3^(x+1)((2/3)^(x+1) + 1))/(3^x((2/3)^x+1))`

Since `3^(x+1) = 3^x*3` yields:

`lim_(x->oo) (3^x*3((2/3)^(x+1) + 1))/(3^x((2/3)^x+1))`

Reducing duplicate factors yields:

`lim_(x->oo) (3((2/3)^(x+1) + 1))/((2/3)^x+1)`

Since `2^x<3^x` for `x>1` yields that `lim_(x->oo)(2/3)^x = 0` , such that:

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

**Hence, evaluating the given limit, yields **`lim_(x->oo)(2^(x+1)+3^(x+1))/(2^x+3^x) = 3.`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes