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?
Best Answer
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 $x\gt 0$.