Wierdly Rational Fractions

1- Lets first simplify the numerator: 

The numerator of the expression is $ \sum_{n=1}^k nx^{n-1} $, which is the derivative with respect to $x $ of the sum $ \sum_{n=1}^k x^n$:
\[ \sum_{n=1}^k x^n = x + x^2 + x^3 + \cdots + x^k = \frac{x(x^k - 1)}{x - 1} = \frac{x^{k+1} - x}{x - 1}\]
Differentiating this expression with respect to $ x$ gives:
\[ \frac{d}{dx} \left( \frac{x^{k+1} - x}{x - 1} \right) = \frac{[(k+1)x^k -1 ](x - 1) - (x^{k+1} - x)}{(x - 1)^2} = \frac{kx^{k+1} - (k+1)x^k +1}{(x - 1)^2}\]

2- Next we simplify the denominator:

The denominator is $ \sum_{n=1}^k nx^{k-n} $, which can be handled by changing the index of summation by letting $m=k-n$
\[ \sum_{n=1}^k nx^{k-n} = \sum_{m=0}^{k-1} (k-m)x^m = k \sum_{m=0}^{k-1} x^m - \sum_{m=0}^{k-1} mx^m \]
The sum $ k\sum_{m=0}^{k-1} x^m $is:
\[ k\sum_{m=0}^{k-1} x^m = k\frac{x^k-1}{ x-1}=k\frac{(x^k-1)(x-1)}{ (x-1)^2}=\frac{kx^{k+1}-kx^k-kx+k}{(x-1)^2} \]
The term $ \sum_{m=0}^{k-1} mx^m $ is derived by differentiating the geometric series sum:
\[ \sum_{m=0}^{k-1} mx^m=x \sum_{m=0}^{k-1} mx^{m-1}=x \frac{d}{dx} \left( \frac{ x^k-1}{ x-1} \right).\]
hence
\[ \sum_{m=0}^{k-1} mx^m= x \frac{d}{dx} \left( \frac{x^k-1}{ x-1} \right) = x \left( \frac{kx^{k-1}(x-1) - (x^k-1)}{(x-1)^2} \right) = \frac{(k-1)x^{k+1}-kx^k+1}{(x-1)^2}. \]
Thus denominator is 
\[ \sum_{n=1}^k nx^{k-n} =\frac{x^{k+1}-kx+k-1}{(x-1)^2}\]

\subsection*{Step 3: Combine and Simplify for the Limit}

Combining these results:
\[ \frac{\sum_{n=1}^{k} nx^{n-1}}{\sum_{n=1}^{k} nx^{k-n}} = \frac{kx^{k+1} - (k+1)x^k +1}{x^{k+1}-kx+k-1} \]\[ \approx \frac{kx^{k+1} - (k+1)x^k }{x^{k+1}} =\frac{kx^{k+1} - (k)x^k-x^k }{x^{k+1}}=k(1-\frac{1}{x})-\frac{1}{x}, \]
as long as $|x|>1$ so we can ignore $-kx+k-1$ when compared to $x^{k+1}$.