A calculus problem by Abdulrahman El Shafei

$\begin{aligned} L & = \lim_{x \to -\infty} \frac {e^x-e^{-x}}{e^x+e^{-x}} & \small \color{#3D99F6} \text{Multiply up and down by }e^x \\ & = \lim_{x \to -\infty} \frac {e^{2x}-1}{e^{2x}+1} \\ & = \frac {0-1}{0+1} = \boxed{-1} \end{aligned}$

Firstly let's look at the Maclaurin expansion of $e^x$ :

$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \displaystyle\sum_{n=0} \frac{x^n}{n!}$ .

Evaluating and canceling out terms $e^x - e^{-x} = 1 - 1 + \frac{x}{1!} + \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^2}{2!} + \cdots = 2(\frac{x}{1!} + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots)$

Since $\sinh{x} = \frac{x}{1!} + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$ , $e^x - e^{-x} = 2\sinh{x}$

Evaluating and canceling out terms $e^x + e^{-x} = 1 + 1 + \frac{x}{1!} - \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^2}{2!} + \cdots = 2(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots)$

Similarly, $\cosh{x} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots$ , $e^x + e^{-x} = 2\cosh{x}$

Therefore $\dfrac { { e }^{ x }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } = \dfrac{2\sinh{x}}{2\cosh{x}} = \dfrac{\sinh{x}}{\cosh{x}}$ and since as $x\to-\infty, \sinh{x}\to \infty, \cosh{x}\to-\infty$ $\dfrac{\infty}{-\infty} = \boxed{-1}$ .