$\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}$$
I tried to use 
$$\lim_{x \to 2} \frac{(x^2)^n-4^n}{x^2-3x+2}$$ but i can't find anything special
 A: Using $a^n - b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots + b^{n-1})$, we get
$$
x^{2n}-4^n = x^{2n}-2^{2n}=(x-2)(x^{2n-1}+\cdots + 2^{2n-1}).
$$
Then divide numerator and denominator through $x-2$, then
\begin{align}
\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}&=\lim_{x\to 2}\frac{(x-2)(x^{2n-1}+\cdots + 2^{2n-1})}{x^2-3x+2}\\
&=\lim_{x\to 2}\frac{x^{2n-1}+\cdots+ 2^{2n-1}}{x-1}\\
&=\frac{2^{2n-1}+2\cdot 2^{2n-2}+\cdots+2^{2n-1}}{1}\\
&=n\cdot 2^{2n}.
\end{align}
A: I think can use $$\lim _{x\to2}\frac{\frac{d}{dx}(x^{2n}-4^n)}{\frac{d}{dx}(x^2-3x+2)}=\lim _{x\to2}\frac{2nx^{2n-1}}{2x-3}=2^{2n}n$$
A: you can see that when u replace x by 2 you get $\frac{0}{0}$ form so l'hopital's rule says when you get a limit of that form you can differentiate both numerator and denominator separately(by assuming they are two different functions)unless indetermined form is removed then replace the value of $x$ and you will get answer in answer of Alexis he used l'hopital's rule
A: L'Hopital
$$\lim\limits_{x\to 2}\frac{2nx^{2n-1}}{2x-3}\overset{x=2}{=}\frac{4^nn}{1}=4^nn$$
