Limit of so many powers

$\begin{aligned} L & = \lim_{x \to \infty} \frac {(x+1)^{10}+(x+2)^{10}+(x+3)^{10}+\cdots +(x+100)^{10}}{x^{10}+10^{10}} & \small \color{#3D99F6} \text{Divide up and down by } x^{10} \\ & = \lim_{x \to \infty} \frac {\left(1+\frac 1x \right)^{10}+\left(1+\frac 2x\right)^{10}+\left(1+\frac 3x\right)^{10}+ \cdots + \left(1+\frac {100}x \right)^{10}}{1+\frac {10^{10}}{x^{10}}} \\ & = \frac {100}1 = 4(25) \end{aligned}$

Therefore, $n = \boxed{25}$ .

The above numerator can be expressed as the following:

$100x^{10} + {10 \choose 1} \cdot (1+2+...+100)x^9 + {10 \choose 2} \cdot (1^2+2^2+...+100^2)x^8 + ... + {10 \choose 9} \cdot (1^9+2^9+...+100^9)x + (1^{10} + 2^{10} + ... + 100^{10})$

using binomial expansions. Since the degree of the numerator polynomial equals the degree of the denominator polynomial, the limit is simply computed to 100 as $x \rightarrow \infty$ and $\boxed{n = 25}.$ above.

Consider a function $y= (x+a)^k$ then the nth derivative of function we have is $y_n = k! (x+a)^{k-n}$ Now $\begin{aligned} &L = \dfrac{1}{x^{10} +10^{10}} \left(\sum_{k=1}^{100} \,(x+k)^{10}\right)\\& L_{10}= \dfrac{1}{9!x}\left(9!\sum_{k=1}^{100}\,(x +k)\right) = \dfrac{1}{x}\left(100x+\sum_{k=1}^{100} k\right) =100 =25\times 4 \end{aligned}$ So $n=25$ .