Euler Formula for Sine Function/Complex Numbers/Proof 1/Lemma 1

Theorem
The function:


 * $\dfrac {\sinh x} x$

is increasing for positive real $x$.

Proof
Let $f \left({x}\right) = \dfrac {\sinh x} x$.

By Quotient Rule for Derivatives and Derivative of Hyperbolic Sine Function:


 * $f' \left({x}\right) = \dfrac {x \cosh x - \sinh x} {x^2}$

From Hyperbolic Tangent Less than X, we have $\tanh x \le x$ for $x \ge 0$.

Since $\cosh x \ge 0$, we can rearrange to get $x \cosh x - \sinh x \ge 0$.

Since $x^2 \ge 0$, we have $f' \left({x}\right) \ge 0$ for $x \ge 0$.

So by Derivative of Monotone Function it follows that $f \left({x}\right)$ is increasing for $x \ge 0$.