Find $\lim_\limits{x\to x_0}{\frac{x f(x_0)-x_0 f(x)}{x-x_0}}$ 
Let $f:\mathbb{R}\mapsto \mathbb{R}$ be a function such that:
  $$\lim_\limits{h\to 0}{\frac{f(x_0+h)-f(x_0)}{h}}=2016$$
Prove the existence of the following limit: $$\lim_\limits{x\to x_0}{\frac{x\cdot f(x_0)-x_0\cdot f(x)}{x-x_0}}$$ and find it only in terms
  of $f(x_0)$ and $x_0$.

I really don't know where to start from. Maybe set $x=x_0+h$ and solve the given limit, but $h$ remains in the final limit expression.
 A: The substitution $x = x_0 + h$ gives
$$
{\frac{x\cdot f(x_0)-x_0\cdot f(x)}{x-x_0}} = {\frac{(x_0+h)\cdot f(x_0)-x_0\cdot f(x_0+h)}{h}} = f(x_0) - x_0 \frac{f(x_0 + h) - f(x_0)}{h}
$$
Now you can compute the limit for $h \to 0$ using your first condition.
A: At first I thought that my simple hint in the comment to question was sufficient. But later looking at OP's comment in one of the answers I thought it is better to give an expanded answer.
I believe OP is somewhat "perplexed" by the form of the question which says "Prove the existence of the following limit". This kind of statement seems to suggest that there are two things to be done here: 1) prove the existence of limit and 2) also evaluate the limit.
The good part here is that in such simple limit problems "proving the existence of limit" goes hand-in-hand with "evaluation of the limit" itself so no extra burden is added.
So we first start with the following theorem of limits:
Theorem: If $\lim_{x \to a}f(x) = L$ exists then $\lim_{h \to 0}f(a + h)$ also exists and is equal to $L$. Conversely if $\lim_{h \to 0}f(a + h) = L$ exists then so does $\lim_{x \to a}f(x)$ and is equal to $L$.
This justifies the use of substitutions of type $x = a + h$ in solving limit problems without the addition burden of showing whether the new expression (after substitution) has a limit or not.
Thus from the given limit $$\lim_{h \to 0}\frac{f(x_{0} + h) - f(x_{0})}{h} = 2016$$ it follows (by using substitution $x = x_{0} + h$) that the following limit $$\lim_{x \to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}}$$ exists and its value is $2016$. We write this as $$\lim_{x \to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}} = 2016\tag{1}$$
Next we need few theorems on "algebra of limits"
Theorem 1: $\lim_{x \to a}k = k$ for any constant $k$.
Theorem 2: If $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exist then $\lim_{x \to a}\{f(x) + g(x)\}$ and $\lim_{x \to a}\{f(x) - g(x)\}$ also exist and $$\lim_{x \to a}\{f(x) \pm g(x)\} = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$$
Theorem 3: If $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exist then $\lim_{x \to a}f(x)g(x)$ also exists and then $$\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x)\cdot\lim_{x \to a}g(x)$$
Now my hint says that $$xf(x_{0}) - x_{0}f(x) = (x - x_{0})f(x_{0}) - x_{0}(f(x) - f(x_{0}))$$ and therefore $$g(x) = \frac{xf(x_{0}) - x_{0}f(x)}{x - x_{0}} = f(x_{0}) - x_{0}\cdot\frac{f(x) - f(x_{0})}{x - x_{0}}$$ Define three new functions $A(x), B(x), C(x)$ as $$A(x) = f(x_{0}), B(x) = x_{0}, C(x) = \frac{f(x) - f(x_{0})}{x - x_{0}}\tag{2}$$ such that $$g(x) = A(x) - B(x)C(x)\tag{3}$$ and note that first two functions $A(x), B(x)$ are constants (independent of $x$) and hence by theorem 1 we have $$\lim_{x \to x_{0}}A(x) = f(x_{0}),\,\lim_{x \to x_{0}}B(x) = x_{0}\tag{4}$$ and from equation $(1)$ above $$\lim_{x \to x_{0}}C(x) = 2016\tag{5}$$ Using theorem 3 we see that $\lim_{x \to x_{0}}B(x)C(x)$ exists and its value is $2016x_{0}$. And then using theorem 2 we see that the $$\lim_{x \to x_{0}}g(x) = \lim_{x \to x_{0}}\{A(x) - B(x)C(x)\}$$ exists
and is equal to $f(x_{0}) - 2016x_{0}$.

Normally we don't need to explain in such detail the existence of limits. We just write the phrase "using algebra of limits we have" and perform the calculations. An examiner should not deduct any marks for this. So a proper answer to the question goes as follows:
Using algebra of limits we have
\begin{align}
L &= \lim_{x \to x_{0}}\frac{xf(x_{0}) - x_{0}f(x)}{x - x_{0}}\notag\\
&= \lim_{x \to x_{0}}\frac{(x - x_{0})f(x_{0}) - x_{0}(f(x) - f(x_{0}))}{x - x_{0}}\notag\\
&= \lim_{x \to x_{0}}\left(f(x_{0}) - x_{0}\cdot\frac{f(x) - f(x_{0})}{x - x_{0}}\right)\notag\\
&= f(x_{0}) - x_{0}\lim_{x \to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}}\notag\\
&= f(x_{0}) - x_{0}\lim_{h \to 0}\frac{f(x_{0} + h) - f(x_{0})}{h}\text{ (putting }x = x_{0} + h)\notag\\
&= f(x_{0}) - 2016x_{0}\notag
\end{align}
The existence of the desired limit is proved via its calculation using rules of "algebra of limits" as shown in the steps above.
A: Given  $$\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}=2016$$ Now, we have $$\lim_{x\to x_0}\frac{xf(x_0)-x_0f(x)}{x-x_0}$$ Let $x=x_0+h\implies h\to 0\ as\ x\to x_0$ $$ =\lim_{h\to 0}\frac{(x_0+h)f(x_0)-x_0f(x_0+h)}{(x_0+h)-x_0}$$ $$=\lim_{h\to 0}\frac{x_0f(x_0)+hf(x_0)-x_0f(x_0+h)}{h}$$ $$=\lim_{h\to 0}\frac{x_0f(x_0)+hf(x_0)-x_0f(x_0+h)}{h}$$ $$=\lim_{h\to 0}\frac{hf(x_0)}{h}+ \lim_{h\to 0}\frac{x_0f(x_0)-x_0f(x_0+h)}{h}$$ $$=f(x_0)-x_0\left( \lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}\right)$$ Now, subtituting the given value, $$=f(x_0)-x_0(2016)$$ $$=f(x_0)-2016x_0$$
