How does one go about $\lim_{y\to x} \frac{x^n-y^n}{x-y}$? $$\lim_{y\to x} \frac{x^n-y^n}{x-y}$$
I think it would not be differentiable because if we used L'Hôpital the denominator would become $0$.
I would like to learn which approach would be better and why: the long-division or by differentiation? 
 A: I think this is easier when renaming the variables. Renaming $x$ to $a$ and $y$ to $x$
$$\lim_{y\to x} \frac{x^n-y^n}{x-y}=\lim_{x\to a} \frac{a^n-x^n}{a-x}$$
Now we can see that L'Hospital gives us:
$$\lim_{x\to a} \frac{a^n-x^n}{a-x} = \lim_{x\to a} \frac{0-nx^{n-1}}{0-1} = nx^{n-1}$$
(The $a^n$ and $a$ are constants and become zero)
But L'Hospital may not be the best choice here.
A: $$\lim _{ y\to x } \frac { x^{ n }-y^{ n } }{ x-y } =\lim _{ y\to x } \frac { \left( x-y \right) \left( { x }^{ n-1 }+{ x }^{ n-2 }y+{ x }^{ n-3 }{ y }^{ 2 }+...+{ y }^{ n-1 } \right)  }{ x-y } =\lim _{ y\to x } \left( { x }^{ n-1 }+{ x }^{ n-2 }y+{ x }^{ n-3 }{ y }^{ 2 }+...+{ y }^{ n-1 } \right) =n{ x }^{ n-1 }$$
A: You can also resort to the binomial theorem and set $h:=x-y$. Then
$$\frac{x^n-y^n}{x-y}=\frac{(y+h)^n-y^n}{h}=\frac{\binom n1y^{n-1}h+\binom n2y^{n-2}h^2+\cdots\binom nnh^n}h,$$ which, after simplification, tends to the first term,
$$ny^{n-1}.$$
A: This is the derivative for $y=x^n$, by definition. Using implicit differentiation and the fact that $\ln(x)=\int_{0}^{x} \frac{1}{t} dt$, and taking logs, we see that
$$\ln(y)= \ln(x^n) \implies \frac{1}{y} \frac{dy}{dx}=n \cdot \frac{1}{x} \implies \frac{dy}{dx}=n \cdot\frac{y}{x}=n \cdot x^{n-1}.$$
Note that this works for any nonzero $n \in \mathbb R$.
A: To find the limit
$\lim_{y\to x} \dfrac{x^n-y^n}{x-y}, \tag 1$
observe that
$x^n - y^n = (x - y) \displaystyle \sum_{i = 0}^{n - 1} x^{n - i - 1} y^i; \tag 2$
so for $y \ne x$,
$\dfrac{x^n - y^n}{x - y} =  \displaystyle \sum_{i = 0}^{n - 1} x^{n - i - 1} y^i; \tag 3$
then
$\lim_{y \to x} \dfrac{x^n - y^n}{x - y} = \displaystyle \sum_{i = 0}^{n - 1} x^{n - i - 1}x^i = \sum_{i = 0}^n x^{n - 1} = nx^{n - 1}. \tag 4$
(4) just demonstrates the fact that the derivative of $x^n$ is $nx^{n - 1}$; however, we may in fact also apply L'hospital's rule to (1), since the limits of the numerator and denominator each exist and are $0$; we are dealing with an indeterminate form of the type $0/0$.  Thus,
$\lim_{y\to x} \dfrac{x^n-y^n}{x-y} = \lim_{y\to x} \dfrac{d(x^n - y^n)/dy}{d(x - y)/dy} = \lim_{y\to x}\dfrac{-ny^{n - 1}}{-1} = nx^{n -1}, \tag 5$
in agreement with (4).
A: By the definition of the derivative:
$$
\begin{align}
\lim_{y\to x}\frac{x^n-y^n}{x-y}
&=\lim_{y\to x}\frac{y^n-x^n}{y-x}\\
&=\frac{\mathrm{d}}{\mathrm{d}x}x^n\\[6pt]
&=nx^{n-1}
\end{align}
$$

As others have pointed out, by polynomial division, we also have
$$
\frac{y^n-x^n}{y-x}=\sum_{k=0}^{n-1}y^{n-k-1}x^k
$$
which, as $y\to x$, tends to $nx^{n-1}$.

Comparing these two methods, the first is more advanced and perhaps a bit more elegant and can be applied to non-integer exponents. However, the second is more basic, with fewer prerequisites. In fact, the latter method is one way to compute the derivative of $x^n$ for $n\in\mathbb{N}$.
