Prove that if:
$\lim_{x\rightarrow\infty}{f(x+1)-f(x)}=L$
than:
$\lim_{x\rightarrow\infty}{\frac{f(x)}{x}}=L$
Assuming $\lim_{x\rightarrow\infty}{f(x+1)-f(x)}=L$ we can choose $X_{\epsilon}$ s.t. $x>X_{\epsilon}$ implies $|f(x+1)-f(x)-L|<\epsilon$ Which means that for $x>X_{\epsilon}$:
$f(X_{\epsilon})+(x-X_{\epsilon})(L-\epsilon) < f(x) < f(X_{\epsilon})+(x-X_{\epsilon})(L+\epsilon) \iff$
$\frac{f(X_{\epsilon})}{x}+(1-\frac{X_{\epsilon}}{x})(L-\epsilon) < \frac{f(x)}{x} < \frac{f(X_{\epsilon})}{x}+(1-\frac{X_{\epsilon}}{x})(L+\epsilon) \iff$
$\frac{f(X_{\epsilon})-X_{\epsilon}(L-\epsilon)}{x}-\epsilon < \frac{f(x)}{x} -L< \frac{f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)}{x}+\epsilon$
Now if we want $|\frac{f(x)}{x}-L|<\alpha$ for $x>X_{\alpha}$ we start by choosing an $\epsilon<\alpha$ and a corresponding $X_{\epsilon}$ s.t. $x>X_{\epsilon}$ implies
$\frac{f(X_{\epsilon})-X_{\epsilon}(L-\epsilon)}{x}-\epsilon < \frac{f(x)}{x}-L < \frac{f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)}{x}+\epsilon$
Now we are going to make sure that for $x>X_{\alpha}$
$-\alpha<\frac{f(X_{\epsilon})-X_{\epsilon}(L-\epsilon)}{x}-\epsilon < \frac{f(x)}{x} -L< \frac{f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)}{x}+\epsilon<\alpha$
First $-\alpha<\frac{f(X_{\epsilon})-X_{\epsilon}(L-\epsilon)}{x}-\epsilon$:
If $f(X_{\epsilon})-X_{\epsilon}(L-\epsilon)\ge0$: $x\ge0$
If $f(X_{\epsilon})-X_{\epsilon}(L-\epsilon)<0$: $x\ge\frac{X_{\epsilon}(L-\epsilon)-f(X_{\epsilon})}{\alpha-\epsilon}$
Now $\frac{f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)}{x}+\epsilon<\alpha$
If $f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)<0$: $x\ge0$
If $f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)\ge0$: $x\ge\frac{f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)}{\alpha+\epsilon}$
So now if we choose $X_{\alpha}=max(X_{\epsilon},\frac{X_{\epsilon}(L-\epsilon)-f(X_{\epsilon})}{\alpha-\epsilon},\frac{f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)}{\alpha-\epsilon})$ Than we know that for $x>X_{\alpha}$:
$-\alpha<\frac{f(X_{\epsilon})-X_{\epsilon}(L-\epsilon)}{x}-\epsilon < \frac{f(x)}{x} -L< \frac{f(X_{\epsilon})-X_{\epsilon}(L+\epsilon)}{x}+\epsilon<\alpha$
And so that $|\frac{f(x)}{x}-L|<\alpha$.
Since we can do this for any $\alpha>0$: $\lim_{x\rightarrow\infty}{\frac{f(x)}{x}}=L$
Now this proof, if even correct, is definitely not very simple/elegant and I feel like this could be done easier. So please if you have a simpler proof share it, and if you see mistakes in this proof point them out. Thanks
