Proving that a function doesn't have a limit given $|a-b| = |f(a)-f(b)|$ for all $a,b\in \mathbb R$ For all $a,b\in \mathbb R$,
$|a-b| = |f(a)-f(b)|$.
Prove that $\;\lim\limits_{a\to \infty}|f(a)|=\infty$
What I was trying to do is show that if we assume that there is a limit, $\;\lim\limits_{a\to \infty}|f(a)|=L\;$ and therefore $\big||f(a)|-L\big|=\big||f(a)|-|L|\big| \leq |f(a)-L| < \epsilon$
Then we set $\epsilon$ to some arbitrary value and try to get a contradiction.
I also tried proving directly but couldn't get much out of it.
Is there a better approach for proving it?
 A: For any $\;a\in\mathbb{R}\;$ it results that
$\begin{align}
|f(a)|&\ge|f(a)-f(0)|-|f(0)|=|a-0|-|f(0)|=\\
&=|a|-|f(0)|\;,
\end{align}$
hence,
$\lim\limits_{a\to\pm\infty}|f(a)|\ge\lim\limits_{a\to\pm\infty}\big(|a|-|f(0)|\big)=+\infty\;,$
consequently,
$\lim\limits_{a\to\pm\infty}|f(a)|=+\infty\;.$
A: Hint By the triangle inequality
$$|f(x)| \geq |f(x)-f(0)|-|f(0)| = |x-0|-|f(0)|=|x|-|f(0)|$$
A: Consider $b=0$.
As $a \to \infty,\ |f(a)| + |f(0)| \geq|f(a)-f(0)| = |a-0| = |a| \to \infty$.
Hence as $a \to \infty,\ |f(a)| \geq |a| - |f(0)| \to \infty.$
A: $$\forall a,b, \quad \frac{f(a)-f(b)}{a-b} = 1 \text{ or }-1.$$
Without loss of generality, take this difference quotient to be "1".  A similar argument will hold when this difference quotient is "-1".  Also, it should be clear that it is one or the other, not both.$^*$
$$f^\prime(x) = 1, \quad \text{everywhere}.$$
Take $f(0)= y_0$, some finite number.
Integrating,
$$f(x) = x+y_0$$
$$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} (x+ y_0) = \infty.$$
$^*$Here is a proof.  If, for $a,b,c$ all not equal, $f(a)-f(b) = a-b$ and $f(b)-f(c)=c-b$ then $f(a)-f(c)= a+c-2b$ so that $a+c-2b=a-c$ and $b=c$ or $a+c-2b=c-a$ and thus $a=b$.  Contradiction.  Thus $f(a)-f(b)=a-b$ everywhere or $f(a)-f(b)=b-a$ everywhere.
