# How do you find the limit of (1+(7/x)+(3/x^2))^x as x approaches infinity?

Feb 26, 2017

See below.

#### Explanation:

Solving $1 + \frac{7}{x} + \frac{3}{x} ^ 2 = 3 \left(\frac{1}{x} + {r}_{1}\right) \left(\frac{1}{x} + {r}_{2}\right)$ we have

$\left(1 + \frac{7}{x} + \frac{3}{x} ^ 2\right) = 3 {r}_{1} {r}_{2} \left({r}_{1} / x + 1\right) \left({r}_{2} / x + 1\right)$ and also

${\left(1 + \frac{7}{x} + \frac{3}{x} ^ 2\right)}^{x} = {\left(3 {r}_{1} {r}_{2}\right)}^{x} {\left({r}_{1} / x + 1\right)}^{x} {\left({r}_{2} / x + 1\right)}^{x}$

but $3 {r}_{1} {r}_{2} = 1$ so

${\lim}_{x \to \infty} {\left(1 + \frac{7}{x} + \frac{3}{x} ^ 2\right)}^{x} = {\lim}_{{y}_{1} \to \infty} {\left(\frac{1}{y} _ 1 + 1\right)}^{{y}_{1} / r 1} {\lim}_{{y}_{2} \to \infty} {\left(\frac{1}{y} _ 2 + 1\right)}^{{y}_{2} / {r}_{2}}$

giving

${\lim}_{x \to \infty} {\left(1 + \frac{7}{x} + \frac{3}{x} ^ 2\right)}^{x} = {e}^{\frac{1}{r} _ 1} {e}^{\frac{1}{r} _ 2} = {e}^{\frac{1}{r} _ 1 + \frac{1}{r} _ 2}$

with ${r}_{1} = - \frac{1}{6} \left(7 + \sqrt{37}\right)$ and ${r}_{2} = - \frac{1}{6} \left(7 - \sqrt{37}\right)$