A "zoom" on the graph of $y=f(x)$ near $(x_0,y_0)$ (with $y_0 = f(x_0))$ with magnification factor $M$ (the same in both $x$ and $y$ directions) is the graph of the function defined by $f(x_0 + x/M) = y_0 + y/M$. Prove that if $f$ is differentiable at $x_0$, then the zoom converges to the straight line through the origin with slope $f'(x_0)$, as $M \to \infty$. What happens to the zoom of $|x|$ near the origin?
I need help with the general case. For the specific case $|x|$ near the origin, we have
$$f(x_0 + x/M) = y_0 + y/M$$
Plugging in for the points $(0,0)$, we have $$f(x/M) = |x/M| = y/M \Longleftrightarrow x/M = y/M \text{ or } -x/M = y/M$$ Multiplying by $M$ on both sides yields $$y = x, y = -x$$ Taking the derivative, we have $y'=1, y' = -1$, so the function isn't differentiable at the origin.
How does the general case work?
I'm getting stuck at the point $M(f(x_0+x/M)-y_0) = y$. I'm having trouble because I don't know $f$ and so I don't quite know how to proceed.
Hint:
\begin{align*} Z_{x_0,M}(f)(x) &=M \left(f\left(x_0+\dfrac{x}{M}\right)-f(x_0)\right)\\ &= \dfrac{f\left(x_0+\dfrac{x}{M}\right)-f(x_0)}{\dfrac{x}{M}}\,x\xrightarrow{M\to \infty}f'(x_0)x, \end{align*}
where the convergence is not uniform in $x$.
On the other hand, we have
$$Z_{0,M}(|\bullet|)(x)=M|x/M|=|x|,$$
so that the absolute value function is fixed under zoom at $0$.
Here are some graphs (for $f:x\mapsto \sin(x)$):
(The link to the interactive graph is: https://www.desmos.com/calculator/kdw5vuaytr)
Finally note that the notion of a tangent cone in metric geometry is closely related; see Why does velocity not have a meaning in a metric space?.
Added: It seems this is Exercise 9 on p.153 of Strichartz' The Way of Analysis, revised edition. Given that Strichartz also wrote a survey on subriemannian geometry, it's almost certain in my eyes that the tangent cone construction was an influence for this exercise.