About limits and successions This is a problem which I am not sure I solved correctly, mainly because there are some passages which are not very rigourous.

Let $\{x_n\} $ be an increasing succession such that $x_n > 0$ and the
  tangent at the point $x_n$ of the function $y = \cos x$ passes through
  the origin. Let $c_n$ be the coefficient of the tangent line at the
  point $x_n$, calculate
$$\lim_{n\to \infty} x_n - n\pi$$ and $$\lim _{n \to \infty} nc_{2n}$$

My attempt
Well the first thing I did was finding that it must be
$\tan x_n = -\frac{1}{x_n}$
So it seems easy to see (from the graphics) from this that $x_n \to \infty$ as $n \to \infty$
This implies than $\lim_{n \to \infty} \tan x_n = 0 \Rightarrow x_n \sim n\pi$
But this is not enough to prove the first limit as it is an insufficient approximation.
So I'm going to say that since $\tan {x_n} \sim x_n - n\pi$, then 
$$x_n = -\frac{1}{\tan x_n} \sim -\frac{1}{x_n - n\pi}$$
Basically I'm identifying $x_n$ with its second order approximation.
from the last equation one easily find $$x_n = \frac{n\pi + \sqrt{n^2\pi^2 - 4}}{2}$$
So $$\lim_{n \to \infty} x_n - n\pi = -\frac{2}{n \pi} = 0$$
For the second one, $$c_{2n} = -\sin(x_{2n}) \sim \sin(-\frac{2}{n \pi}) \sim -\frac{2}{n \pi}$$
So $$\lim _{x \to \infty} nc_{2n} = n \cdot -\frac{2}{n \pi} = -\frac{2}{\pi}$$
 A: I guess there's supposed to be some assumption which ensures that $x_n$ occurs on the $n$th "bump" of the cosine wave.  Otherwise there is nothing I can see in the question as stated to prevent $x_n$ from growing arbitrarily fast, destroying any hope of evaluating $\lim_{n\to\infty} (x_n-n\pi)$.  So I'll just assume that
$$ n\pi - \frac\pi2 \le x_n \le n\pi \tag{$\ast$} $$
(If that needs more justification, let me know.  Edit: see below.)  Having this statement available simplifies things a lot.
For the first part, ($\ast$) shows $x_n\to\infty$; combining that with your finding that $\tan(x_n) = -1/x_n$ yields
$$ |\sin(x_n-n\pi)| = |\sin x_n| = \frac{|\cos x_n|}{x_n} \to 0 $$
which since $x_n-n\pi\in[-\frac\pi2,0]$ (again by ($\ast$)) implies $x_n-n\pi\to 0$.
For the second part, I'd like to write it this way:
$$ nc_{2n} = -n\sin(x_{2n}) = \frac{n}{x_{2n}} \cos(x_{2n}) = \frac{n}{x_{2n}} \cos(\underbrace{x_{2n}-2n\pi}_{\to0}) \to \frac2\pi $$
Edit:  Here's some more about ($\ast$).  As you said, the tangent condition in the problem is equivalent to saying that the $x_n$ are solutions to the equation
$$ \tan x = -\frac1x \tag{$\dagger$} $$
Lemma.  Equation ($\dagger$) has exactly one solution in each interval $I_k = (k\pi-\frac\pi2,k\pi)$, and no other positive solutions.
Proof.  Let $g(x) = \tan x + 1/x$.  First consider intervals of the form $[k\pi,k\pi+\frac\pi2)$, where $k$ is a nonnegative integer.  On such intervals $\tan$ is nonnegative, so $g>0$.  Thus there are no roots in such intervals, as desired.  Next consider intervals $(k\pi-\frac\pi2, k\pi)$ where $k$ is a positive integer.  Since $g$ is continuous on such intervals and $\lim_{x\to (k\pi-\frac\pi2)^+} g(x) = -\infty$ and $\lim_{x\to k\pi^-} g(x) = 1/k\pi > 0$, by IVT the function $g$ has at least one zero in each such interval.  On the other hand, if $g$ had two or more zeroes in any such interval, then by Rolle's theorem $g'$ would have a zero in that interval, which is impossible since $g'(x) = \sec^2 x - 1/x^2 \ge 1 - 1/(\frac\pi2)^2 > 0$.
(If you want a proof without so much calculus, use the fact that the cosine function alternates between being positive and concave and being negative and convex, which I guess can be shown directly from the addition formula.)
From the lemma we can see that the conditions given in the problem are are not sufficient to determine $\lim_{n\to\infty} (x_n-n\pi)$.  Indeed, let the $n$th positive solution of equation ($\dagger$) be denoted $a_n$.  The conditions of the problem — that the sequence $\{x_n\}$ is increasing and its terms are positive solutions of ($\dagger$) — just mean that the sequence $\{x_n\}$ is a subsequence of $\{a_n\}$.  For example, we could have $x_n = a_{n^2}$, for which we'd have $x_n - n\pi \ge n^2\pi - \frac\pi2 - n\pi \to \infty$.
So we need to assume something else about the $x_n$ to solve the problem.  The simplest thing is to assume that $x_n=a_n$.  (That assumption might be more aesthetically pleasing in another form, e.g., we could assume that the sequence $\{x_n\}$ grows as slowly as possible given the other assumptions.)
