Prove $\sinh x > x$ for all $x >0$

Question

I did a proof for $\sinh x > x$ for all $x > 0$. But I am not sure if the proof is mathematically valid.

I started by showing that $\frac{d}{dx} \sinh x = \cosh x$ and that the limit of $\cosh x = 1$ and that $\cosh x > 1$ for all $x > 0$. This means that $\sinh x$ will always grow faster than $x$.

Next i created a function $f(x) = \frac{\sinh x}{x}$ and showed that the limit of that function was also one. I did that using L'Hospital's Rule as so:

$lim_{x \rightarrow 0^+} f(x) = lim_{x \rightarrow 0^+} \frac{\cosh x}{1} = 1$

I then concluded by saying that since for all $x > 0$, $\sinh x$ will always grow faster than $x$ and and that $lim_{x \rightarrow 0^+} f(x) = 1$ that means that $\sinh x > x$ for all $x > 0$.

I would love for anyone to tell me if this is a valid proof, and if it isn't how I could go about making it valid.

EDIT: At my university they hinted that we should use the mean value theorem. But I couldn't see any use for it. Is it necessary?

Answer 1

Here is a less complicated version of your argument. Let $$f(x)=\sinh x-x.$$ Note that $f(0)=0$, and that $f'(x)=\cosh x-1\gt 0$ for $x\gt 0$. Thus $f$ is an increasing function on $[0,\infty)$, and therefore $f(x)\gt 0$ for $x\gt 0$.

Remark: The Mean Value Theorem is implicitly used, since we used the fact that since $f'(x)\gt 0$, $f(x)$ is increasing. The usual proof of that fact uses the MVT. But one could use the MVT explicitly in this case as follows. By the MVT, $$\frac{\sinh x-0}{x-0}=\cosh(c)$$ for some $c$ between $0$ and $x$. But since $\cosh c\gt 1$, we have $\frac{\sinh x}{x}\gt 1$ if $x\gt 0$.

Answer 2

It looks OK. Easier proof using Taylor series: $$\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ... > x.$$

Answer 3

The power series is the key. Recall that $$\mathrm f(x) \sim \mathrm f(0) + \mathrm f'(0)\cdot x + \frac{\mathrm f''(0)}{2!}\cdot x^2 + \frac{\mathrm f'''(0)}{3!}\cdot x^3 + \cdots + \frac{\mathrm f^{(k)}(0)}{k!}\cdot x^k+\cdots$$ Since $\sinh x$ is expressed as a linear combination of $\mathrm e^x$ and $\mathrm e^{-x}$, which are themselves holomorphic functions, we know that $\sinh x$ is holomorphic. It is not hard to show that $$\sinh x = x + \frac{1}{3!}\,x^3 + \frac{1}{5!}\,x^5 + \frac{1}{7!}\,x^7 + \cdots + \frac{1}{(2k+1)!}\,x^{2k+1} + \cdots $$

Clearly $\sinh x > x$ for all $ x > 0$.

Answer 4

Simpler and more general: use this direct consequence of the Mean value theorem:

Let $f$ and $g$ be two differentiable functions on an interval $I$, $x_0\in I$. Suppose for all $x>x_0$ in $I$, we have $f'(x)>g'(x)$. Then, for all $x>x_0$, we have $$f(x)-f(x_0)> g(x)-g(x_0).$$

Answer 5

Since $\sinh'(x) =\cosh(x) $, $\cosh(x) =\frac12(e^x+e^{-x}) =\frac12((e^{x/2}-e^{-x/2})^2+2) > 1 $ for $x > 0$, and $\sinh(0) = 0$, $\sinh(x) =\int_0^x \cosh(t)dt \gt\int_0^x 1dt = x $.