Prove that $\lim_{x \to y} \frac{x^k-y^k}{x-y} = k*y^{k-1}$ for $y\in\mathbb R$ and $k\in\mathbb N$ how can I prove the following:
Prove that
$$
\lim_{x \to y}  \frac{x^k-y^k}{x-y} = ky^{k-1}
$$
for a fixed $y\in\mathbb R$ and $k\in\mathbb N$.
I already tried a lot but somehow I don't come to the correct conclusion
 A: Using
$$x^k -y^k= (x-y)\left(\sum_{i=0}^{k-1} x^{k-i-1}y^i\right)$$
the limit is $$\lim_{x \to y}\sum_{i=0}^{k-1} x^{k-i-1}y^i = \sum_{i=0}^{k-1} y^{k-1}=ky^{k-1}$$
A: The derivative of a function $f(x)$ is
$$
\frac{df}{dx} = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
$$
Let $f(x) = x^k$ and $h = x-y$. Then:
$$
\frac{df}{dx} = \lim_{h \to 0} \frac{(x+h)^k-x^k}{h}
$$
From the binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem) you can see that
$$
\frac{df}{dx} = \lim_{h \to 0} \frac{(x^k + k x^{k-1} h + \cdots )-x^k}{h} = \lim_{h \to 0} \frac{k x^{k-1} h + \cdots}{h}
$$
The terms included in the $\cdots$ involve higher powers of $h$, i.e., $h^2$, $h^3$ and so on. If you divide all by $h$, you will have
$$
\frac{df}{dx} = \lim_{h \to 0} k x^{k-1} + \cdots
$$
If you make $h\to 0$, the terms included in $\cdots$ vanishes, and you have
$$\frac{df}{dx} = k x^{k-1}$$
Then:
$$
\lim_{x \to y} \frac{x^k-y^k}{x-y} = k x^{k-1}
$$
Naturally, you can see that $df/dx = k x^{k-1}$ using derivative rules.
