Question

How do you determine the limit of `5^x/(3^x+2^x)` as x approaches infinity?

Answer 1

Use l'Hopital's Rule and some algebra.

In the original form, l'Hopital's rule doesn't help, but note that:

`5^x/(3^x+2^x)=(5^x/2^x)/(3^x/2^x+1)=(5/2)^x/((3/2)^x+1)`.

The limit as `xrarroo` is of the form `oo/oo`.

L'Hopital tells us to consider the ratio of the derivatives:

`((5/2)^xln(5/2))/((3/2)^xln(3/2)`.

Note that this is simply `(5/2)^x/(3/2)^x` times the positive constant `ln(5/2)/ln(3/2)`.

Furthermore, `(5/2)^x/(3/2)^x=(5/3)^x`.

So, as `xrarroo`, `5^x/(3^x+2^x)` behaves like `((5/2)^xln(5/2))/((3/2)^xln(3/2)` which in turn behaves like `k(5/3)^x` with `k>0`. Which increases without bound (goes to `oo`) as `xrarroo`.

`lim_(xrarroo)5^x/(3^x+2^x)=oo`.

Bonus: `lim_(xrarr-oo)(5/2)^x/((3/2)^x+1)=0/(0+1)` so `lim_(xrarr-oo)5^x/(3^x+2^x)=0`