How to see $\lim_{x \to 0}\frac{1-\cos x}{x^2}=\frac12$ by looking at the graphs of $1-\cos x$ and $x^2$ We know $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ because $\sin x$ behaves like $x$ locally. It can be seen from the following graph:

We can see as $x$ approaches $0$, $\sin x$ behaves more and more like $y = x$ which makes the limit $\lim_{x \to 0}\frac{\sin x}{x}$ equal to $1$.

Now, let's see another limit. This time, we have
$$\lim_{x \to 0}\frac{1-\cos x}{x^2}$$
which is equal to $\frac12$. Here is the graph of $(1 - \cos x)$ and $x^2$:

How is this limit approaching $\frac12$? How to see it approaching $\frac12$ by the graph as in the case of $\lim_{x \to 0}\frac{\sin x}{x}$?
 A: One might ask what you mean when you say $\sin x$ behaves like $x$ locally.
Presumably you mean "near $x=0$,"
but in calculus courses (even fairly advanced ones) such a statement would
likely only mean that the derivative of $\sin x$ is the same as the derivative of $x$ at $x = 0,$ namely, if $f(x) \triangleq \sin x$ then $f'(0) = 1.$
That is, in Leibniz notation,
$\left.\frac{\mathrm d}{\mathrm dx} \sin x \right|_{x=0} = 1.$
This is not really a statement about the ratio of $\sin x$ and $x$;
it's a statement about derivatives.
It happens, however, that you can apply L'Hôpital's rule to your problem:
if $f(x) \triangleq \sin x$ and $g(x) \triangleq x$
then $\lim_{x \to 0} \sin x = \lim_{x \to 0} x = 0,$ therefore
$$ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)}, $$
and since $\lim_{x \to 0} f'(x) = \lim_{x \to 0} g'(x) = 1,$
the limit of the ratio is $1$ as well.
L'Hôpital's rule, however, is not exactly a trivial fact.
The graph is highly suggestive but is not a proof.
And what do you say about
$$ \lim_{x \to 0} \frac{x}{e^x - e^{-x}} ? $$
The graphs of $y = x$ and $y = e^x - e^{-x}$ look like this:

Neither one behaves like the other locally at $x = 0,$ other than passing through the same point. But you can still evaluate the limit, which (by the way) is $\frac 12.$
Yet again, consider the limit
$$ \lim_{x \to 0} \frac{2(1 - \cos x)}{x^2}. $$
The graphs of $y = 2(1 - \cos x)$, $y = x^2$, and (for comparison) $y = 1 - \cos x$
look like this:

It seems to me that $y = x^2$ looks much more like $y = 2(1 - \cos x)$
than like $y = 1 - \cos x$ near $x = 0.$
In fact, $2(1 - \cos x)$ behaves like $x^2$ locally near $x = 0$
in an even stronger way than $\sin x$ behaves like $x$:
not only do the derivatives match, but the second derivatives match as well.
And since the second derivatives are not zero, this means you can apply
L'Hôpital's rule twice in order to solve the limit. The limit is $1.$
Now replace $2(1 - \cos x)$ by $1 - \cos x$, and the value of the ratio is divided by $2$ at every value of $x$ -- and therefore so is the limit.
$$ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac 12. $$
A: It's not so intuitive but, the blue line gets closer and closer to bisecting the space between the green line and the $x$ axis as $x$ approaches $0$. If you take a series of decreasing values of $x$ close to zero, you get a good visual of where it's heading. Say $x = .2, x = .1, x = .01$ and $x = .001$.
This gives $\frac{1-\cos x}{x^2} = .4983,.4996,.499996$ and $.49999996.$
A: If you graph $y=\cos x$ and superimpose the graph of $y=\dfrac{x^2} 2$ you will still see that graph of the quadratic function touches the graph of the cosine function at just one point. But if $y= cx^2$ for some positive number $c$ that is less than $1/2,$ then the parabola will intersect the curve at three points: one at $x=0$ and two at $x=\pm(\text{something}).$
