# How do you evaluate the limit (1+1/sqrtx)^x as x approaches oo?

Sep 10, 2016

${\left(1 + \frac{1}{\sqrt{x}}\right)}^{x} \to \infty$ as $x \to \infty$

#### Explanation:

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

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

$= {e}^{{\lim}_{x \to \infty} x \ln \left(1 + \frac{1}{\sqrt{x}}\right)} \text{ (*)}$

(The previous step is true as ${e}^{x}$ is a continuous function)

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

$= {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln \left(1 + \frac{1}{\sqrt{x}}\right)}{\frac{d}{\mathrm{dx}} \frac{1}{x}}$ (by L'Hopital's rule)

$= {\lim}_{x \to \infty} \frac{\frac{1}{1 + \frac{1}{\sqrt{x}}} \cdot \frac{1}{x \sqrt{x}}}{\frac{1}{x} ^ 2}$

$= {\lim}_{x \to \infty} \frac{x}{\sqrt{x} + 1}$

$= \infty$

Substituting this into $\text{(*)}$:

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

Thus ${\left(1 + \frac{1}{\sqrt{x}}\right)}^{x} \to \infty$ as $x \to \infty$

Sep 10, 2016

$\infty$

#### Explanation:

Another approach using Bernoulli's compounding formula ie

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

So here:
${\lim}_{x \to \infty} {\left(1 + \frac{1}{\sqrt{x}}\right)}^{x}$

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

with $u = \sqrt{x}$

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

$= {\lim}_{u \to \infty} {e}^{u} = \infty$

Sep 10, 2016

$\infty$

#### Explanation:

Using the binomial expansion

${\left(1 + u\right)}^{r} = {\sum}_{k = 0}^{\infty} \left(\begin{matrix}r \\ k\end{matrix}\right) {u}^{k}$

making $y = \sqrt{x}$

$z = {\left(1 + \frac{1}{y}\right)}^{{y}^{2}} = 1 + {y}^{2} \left(\frac{1}{y}\right) + \frac{{y}^{2} \left({y}^{2} - 1\right)}{1 \times 2} {\left(\frac{1}{y}\right)}^{2} + \cdots$

$z = 1 + y + f \left(y\right)$

${\lim}_{y \to \infty} z = \infty$ because ${\lim}_{y \to \infty} f \left(y\right) \ge 0$