$\lim\limits_{n \to \infty } ({X_{n + 1}} - {X_n}) = c$ implies $\lim\limits_{n \to \infty } {{{X_n}} \over n} = c $ 
Let $\{X_n\}$ be a sequence such that:
$$\lim\limits_{n \to \infty } ({X_{n + 1}} - {X_n}) = c$$
Prove that:
$$\lim\limits_{n \to \infty } {{{X_n}} \over n} = c$$

I've tried many approaches here, but not sure how to connect between the two limits.
Any help will be appreciated.
 A: Fix $\epsilon > 0$.
Limsup:
Choose $N$ large so that $X_n - X_{n-1} \le c + \epsilon$ for all $n \ge N$.
Then choose $M$ large enough so that $\frac{x_N}{M} \le \epsilon$.
for $n \ge \max(M,N)$ write 
$$ x_n = x_N + \left( x_{N+1} - x_{N}\right) + \cdots + \left( x_n - x_{n-1} \right)$$
Then $$\frac{x_n}{n} = \frac{x_{N} + \left( x_{N+1} - x_N\right) + \cdots + \left( x_n - x_{n-1} \right)}{n}$$ 
Hence $\frac{x_n}{n} \le \epsilon + \frac{n - N}{n}(c + \epsilon)$. 
Taking $n\rightarrow\infty$ we conclude that $\limsup \frac{x_n}{n} \le c + 2\epsilon$.
So since $\epsilon$ is arbitrary we conclude that $\limsup \frac{x_n}{n} \le c$. 
Liminf: This time choose $N$ large so that  $X_n - X_{n-1} \ge c - \epsilon$ for $n \ge N$. 
Choose $M$ large so that $\frac{x_N}{M} \ge -\epsilon$. Now run through the same argument as before to conclude that $$\frac{x_n}{n} \ge - \epsilon + \frac{n - N}{n} (c - \epsilon)$$
taking $n \rightarrow \infty$ and conclude that $\liminf \frac{x_n}{n} \ge c - 2\epsilon$. since $\epsilon$ was arbitrary we get $\liminf \frac{x_n}{n} \ge c$.
Combining these we conclude that $\lim \frac{x_n}{n} = c$.
A: Let $\log y_{n} = X_{n}$ and then we get $$\lim_{n \to \infty}\log\left(\frac{y_{n + 1}}{y_{n}}\right) = c$$ so that $\lim_{n \to \infty}y_{n + 1}/y_{n} = e^{c} > 0$ and hence $\lim_{n \to \infty} y_{n}^{1/n} = e^{c}$ and taking logs we get $$\lim_{n \to \infty}\frac{X_{n}}{n} = c$$
