Proving geometrically that, for vectors $u$ and $v$ in $\Bbb{R}^2$, $\lim_{x\rightarrow 0}\frac{|u+xv|-|u|}{x}$ exists and is finite

Question

Let $u,v$ be two vectors in $\mathbb{R}^2$ (e.g. $u=(1,2), v=(-1,1)$).

Consider the function $$x\mapsto \frac{|u+xv|-|u|}{x}$$which is defined for $x>0$. I'm trying to prove in a purely geometric manner that $$\lim_{x\rightarrow 0}\frac{|u+xv|-|u|}{x}$$ exists and is finite.

Analytically this is easy (prove that it is a monotonic, bounded function, and hence all one-sided limits exist and are finite), but geometrically (=without resorting to numbers) I was not able to do it. The basic problem is that I cannot plot this function by purely geometric constructions; from such a plot the existence and finiteness would automatically follow .

Can you do it?

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Philosophical remarks:

I'm having a hard time accepting that it's possible to prove this with an analytic argument that lacks any geometric insight - and that it is not possible to also provide a geometric proof.

This is because it's usually possible to have both. An example from calculus where both approaches are possible, is analyzing differentiability of a function at a point $x_0$: The analytic proof there actually has a strong geometric content, since one relies on the geometric plot of the function to see what goes wrong: if it has a kink at $x_0$ or goes to infinity or oscillates, then this information is directly represented and used in the analytic proof. Thus, here the two approach are intimately connected, whereas in my example above they seem to be divorced.

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This is how far I got in my construction (using Geogebra):

Plot the vectors $u$ and $v$:

Then you can trace out, with the green vector arrows, the map $$x\mapsto u+xv$$ (of course this is the line spanned by $v$, translated via $u$, but let's pretend we didn't knew this already from experience):

Now:

Then we obtain a plot of $$x \mapsto \frac{|u+xv|-|u|}{x},$$ and now of course it is easy to "see" that the limit exists.

Note: Of course one could have obtained this plot in Geogebra, by simply starting to measure the various vector lengths and then dividing that number by the relevant $x$ and marking that spot on the coordinate system (which is probably how Geogebra does it internally).
But I would have preferred obtaining this picture by simply doing geometric constructions (e.g. by using those that are available in Geogebra).

Perhaps no such constructions are possible and going about this problem simply by using numbers (which is what the analytical approach in the end boils down to) is all there is. But then I would like a proof (or at least a justification) for why that is the case.
I feel like I will nee to offer bounty for this questions.

Answer 1

I don't know if this can be of help or not, but if you multiply both numerator and denominator of your function by $(|u+xv|+|u|)$, then that function can be rewritten as: $$ f(x)={xv^2+2(u\cdot v)\over|u+xv|+|u|}. $$ From this it immediately follows that $$ \lim_{x\to0}f(x)={(u\cdot v)\over|u|}, $$ which at least has a simple geometric meaning: it is the projection of $v$ on $u$.

Answer 2

Take $D$ as origin, and $ABCD$ as a parallelogram. $CF$ is arc drawn taking $D$ as centre and $|u|$ as radius. Now, $x\rightarrow 0\Rightarrow BC\rightarrow 0$. But, $$\theta <\tan \theta <\frac{BC}{DC}\rightarrow 0$$ But $\theta+\phi$ is constant, equal to angle between $u$ and $v$. Hence, $\phi$ goes to the angle between $u$ and $v$, as $x\rightarrow 0$. Also, $F\rightarrow E$, with $x\rightarrow 0$. Now, $$\frac{|u+xv|-|u|}{x}=|v|\frac{|u+xv|-|u|}{|xv|}$$ $$=|v|\frac{BF}{BC}$$ $$\rightarrow |v| \frac{BE}{BC}$$ $$=|v|\cos\phi$$ $$\rightarrow \frac{v.u}{|u|}$$ because $\phi$ goes to angle between $u$ and $v$ in the limit.

Answer 3

Let $u:=\overrightarrow{OU}$ and $v:=\overrightarrow{OV}$, and $\overrightarrow{OW_t}:=u+tv$ for some scalar $t$ (instead of $x$).

Let $U'$ be on $\overrightarrow{OW_t}$ such that $|OU|=|OU'|$. Translate $W_t$ by $v$ to $W'$. Construct $\bigcirc UU'W'$, and let $Q$ be the point where the extension of $\overrightarrow{OW_t}$ meets that circle. Define $q:=|W_tQ|$.

By the chord-chord aspect of the power of a point theorem for $W_t$ with respect to the circle, we have $$|UW_t||W_tW'| = |U'W_t||W_tQ|\quad\to\quad t|v|^2=(|u+tv|-|u|)q \quad\to\quad \frac{|u+tv|-|u|}{t}=\frac{|v|^2}{q} \tag{1}$$ This gives us a geometric interpretation of the difference quotient.

Now, as $t\to 0$, points $U$, $U'$, $W_t$ coalesce ...

... and $\overline{W_tQ}$ approaches a diameter of the circle:

(The reader is invited to make this argument more precise, but loosely speaking: the limiting circle circumscribes the "flat" $\triangle UU'Q$, which coincides with the diameter.) This in and of itself guarantees that $q$ has a limiting value, so that the difference quotient $(1)$ does, as well. But we can say more.

By Thales' Theorem, the limiting position has $\angle UW'Q$ as a right angle. Moreover, $\angle UOV = \angle QUW' =: \theta$. Therefore, $$\frac{|v|}{q} = \cos\theta = \frac{u\cdot v}{|u||v|} \quad\to\quad \frac{|v|^2}{q}=\frac{u\cdot v}{|u|} \tag{2}$$ Thus, the limiting value of the difference quotient $(1)$ is given by the right-hand side of $(2)$. $\square$