# Question 0df97

Jul 17, 2016

The answer to 4 is ${e}^{-} 2$.

#### Explanation:

The problem is:

${\lim}_{x \to \infty} {\left(\frac{2 x + 2}{2 x + 4}\right)}^{2 x + 2}$

Now this is a difficult problem. The solution lies in very careful pattern recognition. You may recall the definition of $e$:

$e = {\lim}_{u \to \infty} {\left(1 + \frac{1}{u}\right)}^{u} \approx 2.718 \ldots$

If we could rewrite the limit as something close to the definition of $e$, we would have our answer. So, let's try it.

Note that ${\lim}_{x \to \infty} {\left(\frac{2 x + 2}{2 x + 4}\right)}^{2 x + 2}$ is equivalent to:

${\lim}_{x \to \infty} {\left(\frac{2 x + 4 - 2}{2 x + 4}\right)}^{2 x + 2}$

We can split up the fractions like so:

${\lim}_{x \to \infty} {\left(\frac{2 x + 4}{2 x + 4} - \frac{2}{2 x + 4}\right)}^{2 x + 2}$
$= {\lim}_{x \to \infty} {\left(1 - \frac{2}{2 x + 4}\right)}^{2 x + 2}$

We're getting there! Let's factor out a $- 2$ from the top and bottom:

${\lim}_{x \to \infty} {\left(1 - \frac{2}{2 x + 4}\right)}^{2 x + 2}$
$= {\lim}_{x \to \infty} {\left(1 + \frac{\left(- 2\right)}{- 2 \left(- x - 2\right)}\right)}^{2 x + 2}$
$\to {\lim}_{x \to \infty} {\left(1 + \frac{\cancel{- 2}}{\cancel{- 2} \left(- x - 2\right)}\right)}^{2 x + 2}$
$= {\lim}_{x \to \infty} {\left(1 + \frac{1}{- x - 2}\right)}^{2 x + 2}$

Let us apply the substitution $u = - x - 2 \to x = - 2 - u$:

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

=(1+1/u)^(2(-2-u)+2#
$= {\left(1 + \frac{1}{u}\right)}^{- 4 - 2 u + 2}$
$= {\left(1 + \frac{1}{u}\right)}^{- 2 u - 2}$

The properties of exponents say: ${x}^{a + b} = {x}^{a} {x}^{b}$

So ${\lim}_{x \to \infty} {\left(1 + \frac{1}{u}\right)}^{- 2 u - 2}$ is equivalent to:

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

The properties of exponents also say that: ${x}^{a b} = {x}^{{a}^{b}}$

Which means this further reduces to:

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

By definition, ${\lim}_{x \to \infty} {\left(1 + \frac{1}{u}\right)}^{u} = e$; and using direct substitution on the second limit yields:

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

$= \frac{1}{1 + \frac{1}{\infty}} ^ \left(2\right)$

$= \frac{1}{1 + 0} ^ \left(2\right)$

$= \frac{1}{1} ^ \left(2\right) = 1$

So the solution is...
${\lim}_{x \to \infty} {\left(1 + \frac{1}{u}\right)}^{{\left(u\right)}^{- 2}} {\lim}_{x \to \infty} {\left(1 + \frac{1}{u}\right)}^{- 2}$

$= {\left(e\right)}^{-} 2 \left(1\right)$

$= {e}^{-} 2$