# Question #9a058

Sep 19, 2015

$e$

#### Explanation:

${\lim}_{x \to 0} {\left(\frac{3 + x}{3 - 2 x}\right)}^{\frac{1}{x}} = {\lim}_{x \to 0} {\left(\frac{3 - 2 x + 2 x + x}{3 - 2 x}\right)}^{\frac{1}{x}} =$

$= {\lim}_{x \to 0} {\left(1 + \frac{3 x}{3 - 2 x}\right)}^{\frac{1}{x}} = A$

Let $\frac{3 x}{3 - 2 x} = \frac{1}{t}$, then:

$3 t x = 3 - 2 x \implies x \left(3 t + 2\right) = 3 \implies \frac{1}{x} = \frac{3 t + 2}{3} = t + \frac{2}{3}$

It's obvious that when $x \to 0$ then $t \to \infty$.

$A = {\lim}_{t \to \infty} {\left(1 + \frac{1}{t}\right)}^{t + \frac{2}{3}} =$

$= {\lim}_{t \to \infty} {\left(1 + \frac{1}{t}\right)}^{\frac{2}{3}} \cdot {\lim}_{t \to \infty} {\left(1 + \frac{1}{t}\right)}^{t} =$

$= 1 \cdot e = e$

Note:
$t \to \infty \implies \frac{1}{t} \to 0$

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

Note 2:

${\lim}_{x \to 0} {\left(\frac{3 - 2 x + 2 x + x}{3 - 2 x}\right)}^{\frac{1}{x}} =$

${\lim}_{x \to 0} {\left(\frac{3 - 2 x}{3 - 2 x} + \frac{2 x + x}{3 - 2 x}\right)}^{\frac{1}{x}} =$

${\lim}_{x \to 0} {\left(1 + \frac{3 x}{3 - 2 x}\right)}^{\frac{1}{x}}$