Can the trigonometric limit be evaluated by "induction"? Can this limit: $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}$ be evaluated by 'induction', as in here?
My motivation for asking is this lecture note from MIT's Calculus Open Course Ware, where the professor 'proves' this limit for his students by doing this, and it left me flabbergasted as I had to learn it in a not hard, but complicated way.
As a soft follow up question, If the way used to "prove" the limit isn't correct, why the professor has chosen this way? As an HS student, I think that the class would be fully able to understand a complete proof. (Comparing the area.)
The lecture here
 A: 
We see from Figure 2 that as $\theta$ shrinks, the length $\sin(\theta)$ of the segment gets closer and closer to the length $\theta$ of the curved arc. We conclude that as $\theta\to0$, $\dfrac{\sin\theta}{\theta}\to 1$.

This proves nothing at all: saying that the cathetus gets closer and closer to the arc is exactly the same as saying the sought limit is $1$. This fallacy is commonly known as begging the question.
Yes, it might be “intuitively clear”, but one of the aims of mathematics is avoiding intuition, that can lead to mistakes. It's not difficult to find functions that, tested on a single precision calculator, seem to have a certain limit, but actually have another: we don't really know what happens when the angle “shrinks so much that we're not even able to make a drawing”, which is the same issue as with the kind of limit described above.
There are different levels of rigor, of course. Also the proof you advocate, with areas, suffers from similar problems: the very concept of measure of an angle is far from rigorous and so is the concept of area. But, at beginner's level, it may be used for conveying ideas how limits can be computed. With a simple comparison of areas, we deduce that, for $\theta>0$,
$$
\cos\theta\le\frac{\sin\theta}{\theta}\le\frac{1}{\cos\theta} \tag{*}
$$
and also that $\sin\theta\le\theta$. This latter inequality immediately provides continuity of the sine and hence of the cosine (by a translation), so (*) gives our limit by squeezing. Later on, everything can be made fully rigorous by defining the circular functions by their Taylor series.
If one wants to appeal to intuition, the “getting closer and closer” can be better derived from the isochronism of a pendulum (for small amplitudes), with physical experiments. Definitely not, in my opinion, just from looking at a picture.
A: 
Can this limit: $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}$ be evaluated by 'induction', as in here?

Certainly not. Tables of values can only give us a guess, possibly wrong. For example, from the first table below "we have" $\displaystyle\lim_{x\to 0}\sin(\tfrac{1}{x})=0$ and, from the second, $\displaystyle\lim_{x\to 0}\sin(\tfrac{1}{x})=\tfrac{\pi}{4}$.


Why the professor has chosen this way?

Because it works. If a reasoning with pictures works, we call it a "geometric proof" (example). If it doesn't work, we call it a "paradox" (example).
A: The OP should look at Is the proof of $\lim_{\theta\to 0} \frac{\sin \theta}{\theta}=1$ in some high school textbooks circular?
But not I just want to work on a 'warm-up' exercise.
Consider a point $(x, y)$ on the unit circle $x^2 + y^2 = 1$ with both $x$ and $y$ greater than $0$, so that we have a point in the first quadrant. Two line segments are of interest:
The line segment length $S$ between $(x, 0)$ and $(x, y)$. This is the sine of some angle, but that is not the focus here.
The line segment length $A$ between $(1, 0)$ and $(x, y)$. This is an approximation of arc length on the circle as $x$ goes to $1$.
Show that $\frac{S}{A}$ approaches $1$ as $x$ approaches $1$ from the left.
Hint: This is really elementary and all that is required is understanding some facts about continuous functions. 
While working on this problem draw some pictures and think about taking the next step, looking at $\frac{sin(\theta)}{\theta}$ as $\theta$ goes to $0$.
