How does the theorem 2 of algebra of limits contradict in this Q? Q: $\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}$
Solution in my textbook using theorem 2 of algebra of limits I.e $\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}$ .
$\begin{aligned} \lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1} &=\lim _{x \rightarrow 1}\left[\frac{x^{15}-1}{x-1} \div \frac{x^{10}-1}{x-1}\right] \\ &=\lim _{x \rightarrow 1}\left[\frac{x^{15}-1}{x-1}\right] \div \lim _{x \rightarrow 1}\left[\frac{x^{10}-1}{x-1}\right] \\ &=15(1)^{14} \div 10(1)^{9} \text {} \\ &=15 \div 10=\frac{3}{2} \end{aligned}$
I agree with it using the formula but I also thought of thinking how would it be solved from the long way [Inserting x = 1]
We get :
$\frac{(1)^{15}-1}{1-1} \div\left[\frac{(1)^{10}-1}{1-1}\right]$
$\rightarrow \frac{0}{0} \div \frac{0}{0}=0$
My answer is completely different from the solution of my textbook. How is that possible ?
If anyone wants to argue that we cannot put value of x = 1 just like that. There is another example in my textbook according to which , we can.
In all of them , the Q simply says: Find the limit.
1)$\begin{aligned} \lim _{x \rightarrow 2} \frac{x^{2}-4}{x^{3}-4 x^{2}+4 x} &=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^{2}} \\ &=\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0} \end{aligned}$
Here , x is tending to 2. This is a rational function.
EDIT: I will share a few more examples.


*We have $\lim _{x \rightarrow 1} \frac{x^{2}+1}{x+100}=\frac{1^{2}+1}{1+100}=\frac{2}{101}$


*(iii) $\lim _{x \rightarrow-1}\left[1+x+x^{2}+\ldots+x^{10}\right]=1+(-1)+(-1)^{2}+\ldots+(-1)^{10}$

$x^{15}-1=(x-1)(1+x+x^2+...+x^{13}+x^{14}).$
$x^{10}-1=(x-1)(1+x+x^2+...+x^{8}+x^{9}).$
So if $x\ne 1$ then $$\frac {x^{15}-1}{x^{10}-1}=\frac {1+x+x^2+...+x^{13}+x^{14}}{1+x+x^2+...+x^{8}+x^{9}}$$ which obviously converges to $\frac {15}{10}$ as $x\to 1.$
