Find the limits :
$$\lim_{x \to 1} \frac{x^{100}-2x+1}{x^{50}-2x+1}$$
Using the L'Hopital's Theorem :
$$\lim_{x \to 1} \frac{x^{100}-2x+1}{x^{50}-2x+1}=\lim_{x \to 1} \frac{100x^{99}-2}{50x^{49}-2}=\frac{98}{48}$$
Now how without use L'Hopital's Theorem.
On
$$\lim_{x \to 1} \frac{x^{100}-2x+1}{x^{50}-2x+1}=\lim_{x \to 1} \frac{x^{100}-x-x+1}{x^{50}-x-x+1}=$$ $$=\lim_{x \to 1} \frac{(x-1)(x(x^{98}+x^{97}+...+1)-1)}{(x-1)(x(x^{48}+x^{47}+...+1)-1)}=\frac{99-1}{49-1}=\frac{49}{24}.$$
On
As seen from multiple limit problems on this website it is clear that the formula $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1}\tag{1} $$ is not as popular as the limit formula $$\lim_{x\to 0}\frac{\sin x} {x} =1$$ The expression under limit in question can be written as $$\dfrac{\dfrac{x^{100}-1} {x-1} -2} {\dfrac{x^{50}-1}{x-1}-2} $$ and this tends to $(100-2)/(50-2)=49/24$ via formula $(1)$.
Hint: $x^n-2x+1 = (x^n-1)-2(x-1) = (x-1)(x^{n-1}+x^{n-2}+\dots+x+1-2)$.