# How do you evaluate  (1+a/x)^x as x approaches infinity?

Jun 5, 2016

${\lim}_{x \to \infty} {\left(1 + \frac{a}{x}\right)}^{x} = {e}^{a}$

#### Explanation:

Define $y = {\lim}_{x \to \infty} {\left(1 + \frac{a}{x}\right)}^{x}$. Now take the natural log to get $\ln \left(y\right) = {\lim}_{x \to \infty} x \cdot \ln \left(1 + \frac{a}{x}\right)$. Now ignore the left side and focus on the right side. The right side can be rewritten as

${\lim}_{x \to \infty} \frac{\ln \left(1 + \frac{a}{x}\right)}{\frac{1}{x}}$

You can try evaluating this limit by plugging in infinity directly. You'll get $\frac{0}{0}$ which is indeterminate form. So apply L'Hôpital's rule and differentiate the numerator and denominator.

${\lim}_{x \to \infty} \frac{\left(\frac{1}{1 + \frac{a}{x}}\right) \cdot \left(- \frac{a}{{x}^{2}}\right)}{- \frac{1}{{x}^{2}}} = {\lim}_{x \to \infty} \left(\frac{a}{1 + \frac{a}{x}}\right) = a$

So we have evaluated this limit, but remember this limit is equal to $\ln \left(y\right)$. So solve for y to get the final answer.

$y = {\lim}_{x \to \infty} {\left(1 + \frac{a}{x}\right)}^{x} = {e}^{a}$

Jun 5, 2016

${\lim}_{x \to \infty} {\left(1 + \frac{a}{x}\right)}^{x} = {e}^{a}$

#### Explanation:

${\left(1 + \frac{a}{x}\right)}^{x} = {\left({\left(1 + \frac{a}{x}\right)}^{\frac{x}{a}}\right)}^{a}$
${\lim}_{x \to \infty} {\left(1 + \frac{a}{x}\right)}^{x} = {\left({\lim}_{x \to \infty} {\left(1 + \frac{a}{x}\right)}^{\frac{x}{a}}\right)}^{a} = {e}^{a}$