How to evaluate the limit $\lim\limits_{x \to 1} \left(\frac{2}{1-x^2} - \frac{3}{1-x^3}\right) $, and others? $$ \lim_{x \to 1} \left( \frac{2}{1-x^2} - \frac{3}{1-x^3} \right)$$
In my opinion the function is not defined at $ x = 1 $ but somehow when I look at the graph, it's continuous and there is no break. I learned to look for points where my function is not defined due to division by zero.
So my example here is: I should look for the limit as $x \rightarrow 1$ but I don't know how to do this. I'm not allowed to use L'Hospital. I know how to look for limits if my variable ($n$ or $x$) "runs" to infinity.  I know that $\frac{1}{n}$ when $n \to \infty$ is $0$. I just "know" that. And if $n$ would "go" or "run" or "tend" (what is the right way to call it) to $1$, the limit would be $1$ just by "imagining the $n$ as $1$". Or is that the wrong way? I'm still learning, so please let me know the right way to do it.
Back to the example. Is there an approach of calculating the limit? In my book they suggest to replace $x$ with a sequence $ x_n$ and $ x_n \rightarrow 1$ and then let $ n \rightarrow \infty $

Consider the expression
$$\lim_{x \to 1}
\left(
\frac1{1-x}
-
\frac1{1-x}
\right)
$$
This is "obviously" zero,
even though both terms blow up
as $x \to 1$,
because the singularities cancel.
The same thing happens here.
The basic identity needed here is
$1-x^n
=(1-x)(1+x+x^2+...+x^{n-1})
$.
Using this,
$\begin{align}
\frac{2}{1-x^2} - \frac{3}{1-x^3}
&=\frac{2}{(1-x)(1+x)} - \frac{3}{(1-x)(1+x+x^2)}\\
&=\frac1{1-x}\left(\frac{2}{1+x} - \frac{3}{1+x+x^2}\right)\\
&=\frac1{1-x}\left(\frac{2(1+x+x^2)-3(1+x)}{(1+x)(1+x+x^2)}\right)\\
&=\frac1{1-x}\left(\frac{-1-x+2x^2}{(1+x)(1+x+x^2)}\right)\\
&=\frac1{1-x}\left(\frac{(2x+1)(x-1)}{(1+x)(1+x+x^2)}\right)\\
&=\frac{(2x+1)(x-1)}{(1-x)(1+x)(1+x+x^2)}\\
&=\frac{-(2x+1)}{(1+x)(1+x+x^2)} \quad \text{  (except where } x=1)\\
\end{align}
$
This final expression has no problem
being evaluated as $x \to 1$.
Its value is
$\frac{-3}{2\cdot 3}
=-\frac12
$.
Another way to do this
is to let
$x = 1+y$,
so $y \to 0$
is the same as $x \to 1$.
I find it easier to see a limit
with a variable going to $0$,
so I make this kind of transformation
whenever possible.
Since
$1-x^2 
= 1-(1-y)^2
=1-(1-2y+y^2)
=2y-y^2
$
and
$1-x^3 
= 1-(1-y)^3
=1-(1-3y+3y^2-y^3)
=3y-3y^2+y^3
$,
$\begin{align}
\frac{2}{1-x^2} - \frac{3}{1-x^3}
&=\frac{2}{2y-y^2}- \frac{3}{3y-3y^2+y^3}\\
&=\frac1{y}\left(\frac{2}{2-y}- \frac{3}{3-3y+y^2}\right)
\quad \text{  (factoring }y\text{ from each denominator)}\\
&=\frac1{y}\frac{2(3-3y+y^2)-3(2-y)}{(2-y)(3-3y+y^2)}\\
&=\frac1{y}\frac{-3y+2y^2}{(2-y)(3-3y+y^2)}\\
&=\frac{-3+2y}{(2-y)(3-3y+y^2)}\\
\end{align}
$
Again,
this has no problem as $y \to 0$,
and gives $-\frac12$ as before
(as it should).
Note that,
by doing it as $y \to 0$,
we do not need to know
the factorizations
of $1-x^2$ and $1-x^3$.
The fact that $1-x$ divides these
becomes the more obvious fact
that $y$ divides the expressions
for $1-(1-y)^2$
and $1-(1-y)^3$.
