Equivalent definition of differentiation of a real function. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous at $x=x_0$.
If $\lim_{t \rightarrow 0} \frac{f(x_0+at)-f(x_0+bt)}{(a-b)t}$ is converge, $f$ is differentiable at $x=x_0$.
It is true that this proposition is false when $a=-b$ since $f(x)=|x|$ $(: x_0=0)$ is counterexample in this case. I proved a few years ago that if $a \ne -b$ then f is differentiable at $x=x_0$, but I lost the proof..(Actually, I have a dim memory about whether condition $ab<0$ is additionally needed.) I remember the proof is rather tricky (this is reason that I lost the proof), so I want to find better proofs to remember core ideas easily.
 A: *

*If $a=b$, then the conclusion of differentiability is not true.Let's then assume $a \ne b$.  


$$\dots\dots\dots$$
2. If $b=0$ then differntiability holds because 
$\lim_{t \rightarrow 0} \frac{f(x_0 + at)-f(x_0)}{a\space t} = 
\lim_{u \rightarrow 0} \frac{f(x_0 + u)-f(x_0)}{u} $, which is the definition of differentiability.  
$$\dots\dots\dots$$
3. If $a \ne b$, $a \ne 0$, and $b \ne 0$, then conclusion of differentiability does not hold, because, unlike case 2) above, the value $f(x_0)$ is not involved in the assumption, so we cannot conclude continuity of $f$ at $x_0$. As an example, take $a=1$, $b=2$, $x_0=0$, and $f(t)=\sin ( 2\pi \frac{\ln(|t|)}{\ln(2)})$ for $t \ne 0$. Then $f(x_0 + at) - f(x_0 + bt) = 0$ for all $t \ne 0$, therefore the limit exists, but $f$ is not continuous at $x_0=0$, hence $f$ is not differentiable at $x_0$.  
The example $f(t)=\sin ( 2\pi \ln(|t|))$, or any other even function that is discontinuous at $x_0$ shows that $a=-b \ne 0$ does not imply differentiability of $f$ at $x_0$.
$$\dots\dots\dots$$
4. Finally, lets assume $|a| \ne |b|$, $a \ne 0$, $b \ne 0$, and $f$ is continuous at $x_0$. By symmetry of the assumptions in $a$ and $b$, we may assume $|a| < |b|$.   Let $\lambda = \frac{a}{b}$, so that $|\lambda| < 1$. By a change of variable in the limit we have 
$L = \lim_{t \rightarrow 0} \frac{f(x_0 + t)-f(x_0 + \lambda t)}{(1-\lambda) t} $ exists. We can recast this fact as:
$$ f(x_0 + t)-f(x_0 + \lambda t) = L (1-\lambda) t + o(t) \tag{1} $$
for all $t$ near $0$.
Replacing $t$ by $ \lambda ^j t$ in (1) we get:
$$ f(x_0 + \lambda^j t)-f(x_0 + \lambda^{j+1} t) = L (1-\lambda) \lambda^j t + o(\lambda^j t) \tag{2} $$
since the condition $|\lambda| < 1$ ensures that $ \lambda ^j t$ is near $0$.  
Adding (2) for $j=0,1,.., N-1$, after obvious simplifications of the telescoping sum:
$$ f(x_0 + t)-f(x_0 + \lambda^{N} t) = L (1-\lambda^N) t + o((1-\lambda^N) t) \tag{4} .$$
 Now we take limit $N \rightarrow \infty $, and use the continuity of $f$ at $x_0$ to get for all $t$ near $0$:
$$ f(x_0 + t)-f(x_0) = L \space t + o(t) $$
which is differentiability of $f$ at $x_0$.  
