Question

How do you find the limit `(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))` as `x->oo`?

Answer 1

`oo`

Explanation:

The greatest power of `x` here is `1`, and is in the numerator. A handy rule of thumb is that then the numerator will outgrow the denominator, so the limit is `oo`.

Another way to approach this is to factor the greatest power from the numerator and denominator:

`lim_(xrarroo)(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))=lim_(xrarroo)(x(1+x^(-1/2)+x^(-2/3)))/(x^(2/3)(1+x^(-5/12))`

`=lim_(xrarroo)(x^(1/3)(1+x^(-1/2)+x^(-2/3)))/(1+x^(-5/12))`

`=lim_(xrarroo)x^(1/3)*lim_(xrarroo)(1+x^(-1/2)+x^(-2/3))/(1+x^(-5/12))`

Note that for `p>0`, `lim_(xrarroo)x^-p=0`, so this limit becomes:

`=lim_(xrarroo)x^(1/3)*(1+0+0)/(1+0)`

`=lim_(xrarroo)x^(1/3)`

`=oo`

Answer 2

A very slightly different approach.

Explanation:

`lim_(xrarroo)(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))` has initial form, `oo/oo` which is indeterminate.

We will write a quotient that is equivalent, but whose denominator does not go to infinity.

Factor out of both the top and bottom, the greatest power of `x` in the bottom. That is, factor out `x^(2/3)`.

`(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (x+x^(6/12)+x^(4/12))/(x^(8/12)+x^(3/12))`

`= (x^(8/12)(x^(1/3)+1/x^(2/12)+1/x^(4/12)))/(x^(8/12)(1+1/x^(9/12))`

`= (x^(1/2)+1/x^(1/6)+1/x^(1/3))/(1+x^(3/4))`

`lim_(xrarroo)(x^(1/2)+1/x^(1/6)+1/x^(1/3))/(1+x^(3/4)) = (oo+0+0)/(1+0) = oo`