Using L'Hospital's Rule?

$\lim_{x\to2} \dfrac{63x^5-630x^4+2520x^3-5040x^2+5040x-2016}{126x^4-1008x^3+3024x^2-4032x+2016}$

$= \lim_{x\to2} \dfrac{63(x^5-10x^4+40x^3-80x^2+80x-32)}{126(x^4-8x^3+24x^2-32x+16)}$

$= \lim_{x\to2} \dfrac{63(x^5+5(-2)x^4+10(-2)^2x^3+10(-2)^3x^2+5(-2)^4x+(-2)^5)}{126(x^4+4(-2)x^3+6(-2)^2x^2+4(-2)^3x+(-2)^4)}$

$= \lim_{x\to2} \dfrac{63(x-2)^5}{126(x-2)^4}$ (Binomial Theorem)

$= \lim_{x\to2} \dfrac{(x-2)}{2}$

$= \frac{0}{2}$

$= 0$

L'Hospital's Rule is not involved. Ha!

Using L'Hospital's Rule (I simply insist):

$\displaystyle \lim_{x\to2} \dfrac{63x^5 -630x^4+2520x^3-5040x^2+5040x-2016}{126x^4-1008x^3+3024x^2-4032x+2016} \quad\quad\color{#3D99F6}{\left[\dfrac{0}{0}\right]}\\ \stackrel{\text{L'H}}=\displaystyle \lim_{x\to2} \dfrac{315x^4 -2520x^3+7560x^2-10080x+5040}{504x^3-3024x^2+6048x-4032} \quad\quad\quad\;\color{#3D99F6}{\left[\dfrac{0}{0}\right]}\\ \stackrel{\text{L'H}}=\displaystyle \lim_{x\to2} \dfrac{1260x^3 -7560x^2+15120x-10080}{1512x^2-6048x+6048} \quad\quad\quad\quad\quad\quad\;\,\color{#3D99F6}{\left[\dfrac{0}{0}\right]}\\ \stackrel{\text{L'H}}=\displaystyle \lim_{x\to2} \dfrac{3780x^2 -15120x+15120}{3024x-6048} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\;\;\color{#3D99F6}{\left[\dfrac{0}{0}\right]}\\ \stackrel{\text{L'H}}=\displaystyle \lim_{x\to2} \dfrac{7560x -15120}{3024} \\ =\dfrac{7560(2) - 15120}{3024}\\ =\dfrac{15120 - 15120}{3024}\\ =\boxed{0}$

Let the numerator be $p(x)$ and the denominator be $q(x)$ . Use Horner's Method to rewrite them as polynomials in $(x-2)$ :

\(\displaystyle\begin{align} p(x):&& \begin{array}{rrrrrr} 63 & -630 & 2520 & -5040 & 5040 & -2016 \\ \fbox{2} & 126 & -1008 & 3024 & -4032 & 2016 \\ \hline 63 & -504 & 1512 & -2016 & 1008 & \boxed{0:=p_0} \\ \fbox{2} & 126 & -756 & 1512 & -1008 & \\ \hline 63 & -378 & 756 & -504 & \boxed{0=:p_1} & \\ \fbox{2} & 126 & -504 & 504 & & \\ \hline 63 & -252 & 252 & \boxed{0=:p_2} & & \\ \fbox{2} & 126 & -252 & & & \\ \hline 63 & -126 & \boxed{0=:p_3} & & & \\ \fbox{2} & 126 & & & & \\ \hline \boxed{63=:p_5} & \boxed{0=:p_4} & & & & \end{array},&& p(x)=\sum_{k=0}^5p_k(x-2)^k=63(x-2)^5\\\\

q(x):&& \begin{array}{rrrrr} 126 & -1008 & 3024 & -4032 & 2016\\ \fbox{2} & 252 & -1512 & 3024 & -2016\\ \hline 126 & -756 & 1512 & -1008 & \boxed{0 =: q_0}\\ \fbox{2} & 252 & -1008 & 1008 & \\ \hline 126 & -504 & 504 & \boxed{0 =: q_1} &\\ \fbox{2} & 252 & -504 & & \\ \hline 126 & -252 & \boxed{0 =: q_2} & &\\ \fbox{2} & 252 & & & \\ \hline \boxed{126 =: p_4} & \boxed{0 =: q_3} & & & \end{array}, && q(x)=\sum_{k=0}^4 q_k(x-2)^k=126(x-2)^4 \end{align} \)

Putting all together: $\displaystyle \lim_{x\rightarrow 2}\frac{p(x)}{q(x)}=\lim_{x\rightarrow 2}\frac{ 63(x-2)^{\cancel{5}} }{ 126\cancel{(x-2)^4} }=\lim_{x\rightarrow2}\frac{x-2}{2}=\boxed{0}$