Showing $\lim\frac{x_n}{y_n}=\lim\frac{x_{n+1}-x_n}{y_{n+1}-y_n}$ Let $(y_n)$ be an unbounded sequence of natural numbers with $y_{n+1}>y_n\forall n\in\mathbb{N}$. Let $(x_n)$ be another sequence, and suppose that
$$\lim\frac{x_{n+1}-x_n}{y_{n+1}-y_n}$$
exists. Prove that
$$\lim\frac{x_n}{y_n}=\lim\frac{x_{n+1}-x_n}{y_{n+1}-y_n}$$
Theres also a hint that we could use from a previously used question in class which is here: Proving Two Limits Are Equal to Each Other
My issue is that I don't see a correlation between the two.  Could some explain please? 
 A: So it seems that you want to use the following result to prove Stolz-Cesaro theorem:

Let $P_{ij}$, where $P_{ij}$ is a real number and $i, j$ natural number, allow the three statements to hold:
  
  
*
  
*$P_{ij}\ge 0$ for all $i, j$
  
*$\lim_{i\to\infty}P_{ij}=0$ for all $j$
  
*$\sum_{j=1}^i P_{ij} =1$ for all $i$
Let $(x_i)$ be a convergent sequence and let a sequence $(y_i)$ be defined by
$y_i=\sum_{j=1}^i P_{ij}x_j$ 
Then $(y_i)$ is a convergent sequence and $\lim y_i=\lim x_i$

I will add that this is similar to Silverman–Toeplitz theorem. (Well, the assumption $\sup\limits_i \sum\limits_{j=0}^n P_{ij}<+\infty$ is missing, but it seems that you implicitly assume that the sum $y_i$ exists for each convergent sequence $(x_j)$. Silverman-Toeplitz theorem is stronger in the sense that it allows also negative values - which is why absolute value is added in some place - and it gives sufficient and necessary condition. But still it is close to the result you've included here, so it's probably worth mentioning.)

What about proving Stolz-Cesaro theorem in this form:

Theorem. Let us assume that $\lim\limits_{n\to\infty} \frac{a_n}{b_n}=l$, where $b_n\ge0$ and $\sum\limits_{n=1}^\infty b_n=+\infty$. Then 
  $$\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n a_k}{\sum_{k=1}^n b_k}=l.$$

As mentioned here, this is simply a different form of Stolz-Cesaro theorem.
Proof. Let us use the previous result with $x_j=\frac{a_j}{b_j}$ and
$$P_{ij}=
  \begin{cases}
    \frac{b_j}{b_1+\dots+b_i} & i\ge j, \\
    0 & i<j.
  \end{cases}
$$
The we have $$y_i = \frac{a_1}{b_1} \cdot \frac{b_1}{b_1+\dots+b_i} + \frac{a_2}{b_2} \cdot \frac{b_2}{b_1+\dots+b_i} + \dots + \frac{a_i}{b_i} \cdot \frac{b_i}{b_1+\dots+b_i} = \frac{a_1+\dots+a_i}{b_1+\dots+b_i}.$$
So if we show that $P_i$ fulfills the above condition we get that
$$\lim\limits_{i\to\infty} \frac{a_1+\dots+a_i}{b_1+\dots+b_i} = \lim\limits_{i\to\infty} y_i = \lim\limits_{j\to\infty} x_j = \lim\limits_{j\to\infty} \frac{a_j}{b_j}.$$ 

This proof might seem somewhat unnatural on the first sight. Maybe it helps thinking about the case $b_n=1$ (i.e., $y_n=n$) first. 
