# How do you determine the limit of [(n+5)/(n-1)]^n as n approaches oo?

Jan 25, 2017

${\lim}_{n \to \infty} {\left(\frac{n + 5}{n - 1}\right)}^{n} = {e}^{6}$

#### Explanation:

Write the expression as:

${\left[\frac{n + 5}{n - 1}\right]}^{n} = {\left[\frac{n - 1 + 6}{n - 1}\right]}^{n} = {\left(1 + \frac{6}{n - 1}\right)}^{n}$

Now take the logarithm:

$\ln {\left(\frac{n + 5}{n - 1}\right)}^{n} = n \ln \left(1 + \frac{6}{n - 1}\right)$

and divide and multiply by $\frac{6}{n - 1}$:

$\ln {\left(\frac{n + 5}{n - 1}\right)}^{n} = 6 \left(\frac{n}{n - 1}\right) \left(\frac{\ln \left(1 + \frac{6}{n - 1}\right)}{\frac{6}{n - 1}}\right)$

Now consider the succession: ${b}_{n} = \frac{6}{n - 1}$. Clearly:

${\lim}_{n \to \infty} {b}_{n} = 0$

${\lim}_{n \to \infty} \frac{\ln \left(1 + \frac{6}{n - 1}\right)}{\frac{6}{n - 1}} = {\lim}_{x \to 0} \ln \frac{1 + x}{x} = 1$

As:

${\lim}_{n \to \infty} \frac{n}{n - 1} = 1$

as well, we have then:

${\lim}_{n \to \infty} \ln {\left(\frac{n + 5}{n - 1}\right)}^{n} = 6$

Now, using the continuity of the exponential function:

${\lim}_{n \to \infty} {\left(\frac{n + 5}{n - 1}\right)}^{n} = {\lim}_{n \to \infty} {e}^{\ln {\left(\frac{n + 5}{n - 1}\right)}^{n}} = {e}^{{\lim}_{n \to \infty} \ln {\left(\frac{n + 5}{n - 1}\right)}^{n}} = {e}^{6}$