Let ${x_{n}}$ be sequence of real numbers such that $\lim_{n\to\infty}(x_{n+1}-x_{n})=c$ Let ${x_{n}}$ be sequence of real numbers such that $\lim_{n\to\infty}(x_{n+1}-x_{n})=c$ ,where c is a positive real number .Then the sequence ${\dfrac{x_{n}}{n}}$ 
A) is not bounded
B) is bounded bu no convergent
C) converges to c
D) converges to 0
how to define $x_{n}$?
if $x_{n}=n$ then $(x_{n+1}-x_{n}) =1$ and  $\lim_{n\to\infty}(x_{n+1}-x_{n})=1=c$ and $(\lim_{n\to\infty}{\dfrac{x_{n}}{n}})=1=c$ and
if $x_{n}=\dfrac{1}{n}$ then $(x_{n+1}-x_{n}) =(\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2})$ and  $\lim_{n\to\infty}(x_{n+1}-x_{n})=0=c$ and $(\lim_{n\to\infty}{\dfrac{x_{n}}{n}})=0=c$ 
i.e. $\lim_{n\to\infty}(x_{n})=c$
 A: Since
$$
  \lim_{n \to \infty} (x_{n + 1} - x_n) = c,
$$
we know that for every $\varepsilon > 0$ there exists $N > 0$ such that
$$
  c - \varepsilon < x_{n + 1} - x_n < c + \varepsilon
$$
for all $n \geq N$.
We thus have that for every $m > N$ that
$$
  x_N + (m - N)(c - \epsilon) < x_m < x_N + (m - N)(c + \varepsilon)
$$
or equivalently
$$
  \frac{x_N}{m} + \left(1 - \frac{N}{m}\right)(c - \varepsilon) < \frac{x_m}{m} < \frac{x_N}{m} + \left(1 - \frac{N}{m}\right)(c + \varepsilon).
$$
We have that
$$
  \lim_{m \to \infty} \frac{x_N}{m} + \left(1 - \frac{N}{m}\right)(c - \varepsilon) = c - \varepsilon
$$
and so there exists $M_1 > 0$ such that
$$
  c - 2\varepsilon < \frac{x_N}{m} + \left(1 - \frac{N}{m}\right)(c - \varepsilon)
$$
for all $m \geq M_1$. Similarly, there exists $M_2 > 0$ such that
$$
  \frac{x_N}{m} + \left(1 - \frac{N}{m}\right)(c + \varepsilon) < c + 2\varepsilon
$$
for all $m \geq M_2$. Thus we have that
$$
  c - 2\varepsilon < \frac{x_m}{m} < c + 2\varepsilon
$$
for all $m > M = \max(M_1, M_2, N)$.
Since such an $M$ exists for every $\varepsilon > 0$, we have that
$$
  \lim_{m \to \infty} \frac{x_m}{m} = c.
$$
