# How do you determine the limit of (7(e^(5n))+5n)^(1/n) as n approaches infinity?

Nov 21, 2016

${\lim}_{n \rightarrow \infty} {\left(7 {e}^{5 n} + 5 n\right)}^{\frac{1}{n}} = {e}^{5}$

#### Explanation:

Let

$L = {\lim}_{n \rightarrow \infty} {\left(7 {e}^{5 n} + 5 n\right)}^{\frac{1}{n}}$

Taking the natural logarithm of both sides gives

$\ln \left(L\right) = \ln \left({\lim}_{n \rightarrow \infty} {\left(7 {e}^{5 n} + 5 n\right)}^{\frac{1}{n}}\right)$

Since $\ln \left(x\right)$ is a continuous function this can become

$\ln \left(L\right) = {\lim}_{n \rightarrow \infty} \ln \left({\left(7 {e}^{5 n} + 5 n\right)}^{\frac{1}{n}}\right)$

Rewriting using logarithm rules,

$\ln \left(L\right) = {\lim}_{n \rightarrow \infty} \frac{1}{n} \ln \left(7 {e}^{5 n} + 5 n\right)$

$\ln \left(L\right) = {\lim}_{n \rightarrow \infty} \ln \frac{7 {e}^{5 n} + 5 n}{n}$

The limit is in the indeterminate form $\frac{\infty}{\infty}$ so L'Hospital's rule applies, yielding

$\ln \left(L\right) = {\lim}_{n \rightarrow \infty} \frac{\frac{d}{\mathrm{dn}} \ln \left(7 {e}^{5 n} + 5 n\right)}{\frac{d}{\mathrm{dn}} n}$

$\ln \left(L\right) = {\lim}_{n \rightarrow \infty} \frac{\frac{35 {e}^{5 n} + 5}{7 {e}^{5 n} + 5 n}}{1}$

$\ln \left(L\right) = {\lim}_{n \rightarrow \infty} \frac{35 {e}^{5 n} + 5}{7 {e}^{5 n} + 5 n}$

We see that ${e}^{5 n}$ is the overpowering term in the limit here, so the limit as $n \rightarrow \infty$ will be $\frac{35 {e}^{5 n}}{7 {e}^{5 n}} = 5$. If you are dissatisfied with this explanation, reapplying L'Hospital's rule for two more iterations will get you to the same answer of $5$.

$\ln \left(L\right) = 5$

Rearranging to solve for the limit shows

$L = {e}^{5}$