A question about the proof of the limit of a function at a point. I wish to investigate the statement: $\lim_\limits{x\rightarrow \frac{\pi}{3}}\cos(x) = \frac{1}{2}$. Consider $\epsilon>0$ given to be used as the defining property of an open interval about $y=\frac{1}{2}$.
I want to construct an interval centered at $x=\frac{\pi}{3}$ that will map through the function $\cos(x)$ and be a proper subset of the vertical interval centered at $\frac{1}{2}$.
An initial attempt was the following calculation(using the familiar $\delta$ to define an interval about $x=\frac{\pi}{3}$):
$
\begin{align}
2\epsilon &= \cos(\frac{\pi}{3}-\delta)-\cos(\frac{\pi}{3}+\delta) = \sqrt{3}\sin(\delta)\Rightarrow\delta = \arcsin(\frac{2\epsilon}{\sqrt{3}})
\end{align}
$
Now let $x\in\left(\frac{\pi}{3}-\delta,\frac{\pi}{3}+\delta\right)$ then $f(x)\in\left(\frac{1}{2}+\epsilon,\frac{1}{2}-\epsilon\right)$.
However, this fails to be true.
For example, if $\epsilon=.1$ then the $\epsilon-$interval is $(.4,.6)$ but $\cos(\frac{\pi}{3}+\delta)$ is $.3966554809$ clearly out of bounds.
Personally, I like the intuitive nature of relating these intervals through their inverses and would like to modify my first attempt to achieve a rigorous result. If we treat each side of the limit point separately by using $\delta_1$ and $\delta_2$ and then choosing the minimum of these I am thinking it will always work and thus provide an alternative yet similar proof technique to the formal definition of limit at least for those functions where it is possible to explicitly find an inverse. What do you think? Has anyone ran across a discussion like this in a calculus text?
$\bf{Further\:analysis}$:
Symmetry must be the issue for the $\delta$ that resulted above. Generalizing a bit, let $\delta_1$ represent the length to the left of the limit point $a$ and similarly $\delta_2$ for the right. Then:
$
\begin{align}
\delta_1 &= \left|a -f^{-1}(L+\epsilon)\right|\\
\delta_2 &= \left|a-f^{-1}(L-\epsilon)\right|\\
\end{align}
$
Choosing $\delta = \min\left\{\delta_1,\delta_2\right\}$ will always work if you can find $f^{-1}$.      
 A: The idea of using inverses is not correct. The fact that an inverse exists with the desired properties is dependent on the fact that the original function is continuous. And the fact that the function is continuous gives you the limit in question. Hence trying to use the inverse is a circular approach.
Moreover this approach smacks of solving the inequality $|f(x) - L|<\epsilon $. The implication in the definition of limit is one way whereas solving inequalities via the use of inverses leads to a two way implication. And naturally it involves more work to establish a two way implication compared to establishing a one way implication.
In general it is far simpler to find a suitable (and simple) function $g(x) $ such that $|f(x) - L|<g(x) $ and then try to ensure the inequality $g(x) <\epsilon$. In the current question one is lucky to have the inequality $$|\cos a-\cos b|\leq |a-b|$$ so that we have $$|\cos x-\cos (\pi/3)|\leq |x-(\pi/3)|$$ and then it is obvious that setting $\delta=\epsilon$ does the job.

Surprisingly many textbooks have overlooked this problem with the use of inverses in proving limit via definition. I read about this problem in Spivak's solution manual to his Calculus where Spivak presents the solution first using inverses (function in question is $f(x) =x^{2}$ and $\delta$ is expressed using some square roots and $\epsilon $) and then mentions that the solution is wrong because of the circularity involved. If you have a copy of Spivak's Calculus do have a look at problem 41, page 100, Chapter 5 Limits. 
A: You have $\cos(a+b)=\cos a \cos b - \sin a \sin b$-
Now choose some  small enough $\epsilon >0$ and take a look at the interval $[\frac {\pi}{3} - \epsilon,\frac {\pi}{3} + \epsilon]$.
Our $\cos$ maps this interval into $[\cos \frac {\pi}{3} \cos \epsilon + \sin \frac {\pi}{3} \sin \epsilon, \cos \frac {\pi}{3} \cos \epsilon - \sin \frac {\pi}{3} \sin \epsilon]$.
By taking $\epsilon \to 0$ both intervals shrink and $\frac {\pi}{3}$ is mapped onto $\cos \frac {\pi}{3}$, which proves that $\cos $ is continuous at $\frac {\pi}{3}$, so limit $\lim_{x \to \frac{\pi}{3}} \cos x$ exists, but to prove that it equals $\frac {1}{2}$ we must use some properties of $\cos$, whether geometrical or analytical.
This can also be done in a much more rigorous and non-circular way.
A: You were asked to "investigate" $\lim_{x\to \pi/2} cos(x)= \frac{1}{2}$.  You realize this not true don't you?  That is the reason what you get is "clearly out of bounds".  Yes, you can "treat each side of the limit point separately" since you could always take your "$\delta$" to be the smaller of "$\delta_1$" and "$\delta_2$", if you were able to find those.
A: We have that
$$|\cos x-\cos y|=2\left|\sin\dfrac{x-y}{2}\sin\dfrac{x+y}{2}\right|\le 2\left|\sin\dfrac{x-y}{2}\right|.$$ Now, it easy to show that $$|\sin t|\le t.$$ So, we have that
$$|\cos x-\cos \dfrac {\pi}{ 3}|\le \left|x-\dfrac {\pi}{ 3}\right|.$$ So, given $\epsilon>0$ it is enough to consider $\delta=\epsilon.$
