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

Mar 5, 2015

Use l'Hopital's Rule and some algebra.

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

${5}^{x} / \left({3}^{x} + {2}^{x}\right) = \frac{{5}^{x} / {2}^{x}}{{3}^{x} / {2}^{x} + 1} = {\left(\frac{5}{2}\right)}^{x} / \left({\left(\frac{3}{2}\right)}^{x} + 1\right)$.

The limit as $x \rightarrow \infty$ is of the form $\frac{\infty}{\infty}$.

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 ${\left(\frac{5}{2}\right)}^{x} / {\left(\frac{3}{2}\right)}^{x}$ times the positive constant $\ln \frac{\frac{5}{2}}{\ln} \left(\frac{3}{2}\right)$.

Furthermore, ${\left(\frac{5}{2}\right)}^{x} / {\left(\frac{3}{2}\right)}^{x} = {\left(\frac{5}{3}\right)}^{x}$.

So, as $x \rightarrow \infty$, ${5}^{x} / \left({3}^{x} + {2}^{x}\right)$ behaves like ((5/2)^xln(5/2))/((3/2)^xln(3/2) which in turn behaves like $k {\left(\frac{5}{3}\right)}^{x}$ with $k > 0$. Which increases without bound (goes to $\infty$) as $x \rightarrow \infty$.

${\lim}_{x \rightarrow \infty} {5}^{x} / \left({3}^{x} + {2}^{x}\right) = \infty$.

Bonus: ${\lim}_{x \rightarrow - \infty} {\left(\frac{5}{2}\right)}^{x} / \left({\left(\frac{3}{2}\right)}^{x} + 1\right) = \frac{0}{0 + 1}$ so ${\lim}_{x \rightarrow - \infty} {5}^{x} / \left({3}^{x} + {2}^{x}\right) = 0$