On finding a limit by dividing by the highest exponent

Question

Sometimes it’s easy to divide by the highest exponent to find a limit.

$${{{n^3} + 4{n^2}} \over {\root 3 \of n + \root 4 \of {{n^3}} }}$$

So, in the above example (which I just made up; there’s nothing special about it), you should divide by $n^3$. How do you handle the denominator which involves roots?

Answer 1

You can use the same method for fraction exponent:

$${{{n^3} + 4{n^2}} \over {\root 3 \of n + \root 4 \of {{n^3}} }}\sim_\infty\frac{n^3}{n^{3/4}}\ \text{since} \ n^{1/3}=_\infty o(n^{3/4})$$

Answer 2

Explanation of Exponential Division.

The cube root of $n$ can be written as $$n^{\frac{1}{3}}$$

So it'd be $$\frac{n^{\frac{1}{3}}}{n^3}$$ which is simply $$n^{(\frac{1}{3} - 3)} = \frac{1}{n^{\frac{2}{3}}} = n^{-\frac{2}{3}}$$