Cfrgtkky

try, without L Hopital

$$lim _{xto :y}left(frac{x^{n+1}-y^{n+1}}{x-y}right)=y^nleft(n+1right)$$
try using simple algebra, can not be derived, using only alegebraic tricks.

$$lim_{x to y} frac{x^{n+1}-y^{n+1}}{x-y}= lim_{x to y}( x^n+x^{n-1}y+x^{n-2}y^2+cdots+y^{n})$$ $$hspace{2cm}=y^{n}+y^{n}+cdots+y^{n}=(n+1)y^{n}$$

Hint

Make life easier : let $x=y+t$ to make
$$lim _{xto y}left(frac{x^{n+1}-y^{n+1}}{x-y}right)=lim _{tto 0}left(frac{(y+t)^{n+1}-y^{n+1}}{t}right)$$ and use the binomial theeorem or Taylor series since
$$(y+t)^{n+1}=y^{n+1}+(n+1) t y^n+frac{1}{2} n (n+1) t^2 y^{n-1}+Oleft(t^3right)$$

Using
$$frac{x^{n+1} - y^{n+1}}{x-y} = sum_{k=0}^{n} x^{n-k} , y^{k}$$
then
begin{align}
lim_{x to y} , frac{x^{n+1} - y^{n+1}}{x-y} &= lim_{x to y} sum_{k=0}^{n} x^{n-k} , y^{k} \
&= sum_{k=0}^{n} y^n hspace{5mm} text{sum of $n+1$ terms} \
&= (n+1) , y^n.
end{align}

Let
$$ t=x-y $$

$$ Rightarrow
lim_{xrightarrow y} = lim_{trightarrow 0} $$

Now,

$$ left(xright)^{n+1} =left[left(x-yright)+yright]^{n+1}
=left(t+yright)^{n+1} $$

Expanding using Binomial theorem,
$$ left(t+yright)^{n+1} = c_0^{n+1}t^{0}y^{n+1} + c_1^{n+1}t^{1}y^{n} + cdots + c_{n+1}^{n+1}t^{n+1}y^{0 } $$

$$ Rightarrow left(t+yright)^{n+1} = y^{n+1}+ tleft[ c_1^{n+1}t^{0}y^{n} + c_2^{n+1}t^{1}y^{n-1} +
cdots + c_{n+1}^{n+1}t^{n}y^{0 } right] $$
Rearranging,
$$ frac{ left(t+yright)^{n+1} - y^{n+1} } {t} = c_1^{n+1}t^{0}y^{n} + c_2^{n+1}t^{1}y^{n-1} + cdots + c_{n+1}^{n+1}t^{n}y^{0 } $$

$$ Rightarrow lim_{trightarrow 0} left[ frac{ left(t+yright)^{n+1} - y^{n+1} } {t} right] = lim_{trightarrow 0} left[ c_1^{n+1}t^{0}y^{n} + c_2^{n+1}t^{1}y^{n-1} + cdots + c_{n+1}^{n+1}t^{n}y^{0 } right] $$

$$ Rightarrow lim_{trightarrow 0} left[ frac{ left(t+yright)^{n+1} - y^{n+1} } {t} right] = left(n+1right)y^{n} $$
Substitute the value of t,
$$ Rightarrow lim_{xrightarrow y} left[ frac{ x^{n+1} - y^{n+1} } {x-y} right] = left(n+1right)y^{n} $$

Hence proved!