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) $\lim _{x \rightarrow-1}\left[1+x+x^{2}+\ldots+x^{10}\right]=1+(-1)+(-1)^{2}+\ldots+(-1)^{10}$
 A: The solution you used, that is,

\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}

is correct. But the part where you replaced $x$ with $1$ is wrong.

Why is it that some limits can be solved by replacing $x$ with $a$, but some cannot? This is because of continuity. That is, some functions are "good" that solving for the limit as $x$ approaches $a$ is the same as evaluating the function at $x = a$. These functions are:

*

*Polynomial functions, for all real $x$


*Rational functions, for all $x$ in the domain


*Radical functions, for all $x$ in the domain

*

*For $\sqrt[n]{x}$ and odd $n$, $x$ can be any real number

*For $\sqrt[n]{x}$ and even $n$, $x$ can be any nonnegative real number



*Trigonometric functions, for all $x$ in the domain

*

*For $\sin x$ and $\cos x$, $x$ is a real number

*For $\tan x$ and $\sec x$, $x$ is a real number and $\cos x \neq 0$.

*For $\cot x$ and $\csc x$, $x$ is a real number and $\sin x \neq 0$.



*Exponential functions, for all real $x$


*Logarithmic functions, for all $x$ in the domain


*Any combination (addition, multiplication) of these functions
In summary, as long as the value of $a$ is in the domain of the function which is a combination of these through addition (including) and multiplication (including division, removing the zeroes of the denominator from the domain), we can substitute it.

Why is the limit not the same when I substituted $x = 1$ to $\frac{x^{15} - 1}{x^{10} - 1}$ and when $x$ approached $1$?
This is because $x = 1$ is not in the domain. You'll soon know what functions are "good" once continuity shows up.

See this page for more information.
A: Let,
$$f(x) = \frac {x^{15}-1}{x^{10}-1}$$
Well, I agree with you! that the value of function $f(x = 1)$ is $ \frac {1^{15}-1}{1^{10}-1}$
But,
The value of limit of function $f(x)$ as $x \to 1$ is $\frac 32$

*

*Limit of function: This means you take the value of the function as $x\to a$ doesn't mean $x = a $ it means it's very very very close to $a$ it's never $a$


*Try this: $f(x = 0.99999) = \frac {(0.99999)^{15} - 1}{(0.99999)^{10} - 1} \approx \frac 32$

A: $x^{15}-1=(x-1)(1+x+x^2+...+x^{13}+x^{14}).$
$x^{10}-1=(x-1)(1+x+x^2+...+x^{8}+x^{9}).$
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 $x\to 1.$
