Here's a more general situation:
THM Let $\langle a_n\rangle$ be any sequence of real numbers and suppose that $\langle b_n\rangle $ is a sequence of positive numbers such that $b_n$ is strictly monotone increasing to $\infty$. Then $\liminf_{n\to\infty}\frac{a_n}{b_n}\geq \liminf_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$ $\limsup_{n\to\infty}\frac{a_n}{b_n}\leq \limsup_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$
PROOF We prove the case for $\liminf$; the $\limsup$ case is analogous. Take $\alpha <\liminf_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$
Then there exists $N$ such that for each $k\geq 0$ we have $\alpha <\frac{a_{N+k}-a_{N+k-1}}{b_{N+k}-b_{N+k-1}}$ Since $b_{n+1}>b_n$, we have for $k\geq 0$ that $\alpha \left( {{b_{N + k}} - {b_{N + k - 1}}} \right) < {a_{N + k}} - {a_{N + k - 1}}$
Thus, for any $m\geq 0$, $\eqalign{ \alpha \sum\limits_{k = 0}^m {\left( {{b_{N + k}} - {b_{N + k - 1}}} \right)} & < \sum\limits_{k = 0}^m {\left( {{a_{N + k}} - {a_{N + k - 1}}} \right)} \cr \alpha \left( {{b_{N + m}} - {b_{N - 1}}} \right) &< {a_{N + m}} - {a_{N - 1}} \cr} $
It follows that $\alpha \left( {1 - \frac{{{b_{N - 1}}}}{{{b_{N + m}}}}} \right) < \frac{{{a_{N + m}}}}{{{b_{N + m}}}} - \frac{{{a_{N - 1}}}}{{{b_{N + m}}}}$ and taking $m\to\infty$ $\alpha \leq \mathop {\lim \inf }\limits_{m \to \infty } \frac{{{a_m}}}{{{b_m}}}$
It follows that, for each $\alpha <\liminf\limits_{n\to\infty}\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$ we have $\alpha \leq \liminf\limits_{m \to \infty } \dfrac{{{a_m}}}{{{b_m}}}$, which means $\mathop {\liminf }\limits_{n \to \infty } \dfrac{{{a_{n + 1}} - {a_n}}}{{{b_{n + 1}} - {b_n}}} \leq \liminf\limits_{m\to\infty} \frac{{{a_m}}}{{{b_m}}}$
COR Let $\langle a_n\rangle$ and $\langle b_n\rangle$ be as before. Then if $\ell=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$ exists, so does $\ell'=\lim_{n\to\infty}\frac{a_n}{b_n}$ and $\ell=\ell'$
COR Let $x_n$ be any sequence. If $\lim_{n\to\infty} x_n=\ell$ then $\lim_{n\to\infty}\frac 1 n \sum_{k=1}^n x_k=\ell$
P By the first corollary with $b_n=n$ and $a_n=\sum_{k=1}^n x_k$, we have $\mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 1}^{n + 1} {{a_k}} - \sum\limits_{k = 1}^n {{a_k}} }}{{n + 1 - n}} = \mathop {\lim }\limits_{n \to \infty } {x_{n + 1}} = \mathop {\lim }\limits_{n \to \infty } {x_n}$
which means $\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} = \mathop {\lim }\limits_{n \to \infty } {x_n}$