Evaluating $\lim_{x\to 1}\frac{x^3+x-2}{x^3-x^2-x+1}$

Question

$$\lim_{x\to 1}\frac{x^3+x-2}{x^3-x^2-x+1}$$

In the above question, I tried to solve using factorization. Since putting $x$ as $1$ gives an indeterminate form, therefore, $(x-1)$ is a factor of both numerator and denominator.

After the factors cancel out and putting the limit, what I get is $0$ in the denominator. I haven't thought of the possibility of "infinite" as the answer though.

I tried L'Hospital's rule too, same result. :(

What's wrong with an "infinite" (more formally, "undefined") answer? If you plot the function, you'll see an asymptote at $x=1$. — Blue, Jul 26 '23 at 10:10
After simplications, let $x=y+1$ to get $\frac{y^2+3 y+4}{y (y+2)}$ and use the long division. — Claude Leibovici, Jul 26 '23 at 10:55
@ShrishBhargav, could you read my answer? Tell me if you like it. — Angelo, Jul 26 '23 at 11:50

score 4 · Accepted Answer · answered Jul 26 '23 at 11:44

There does not exist the limit $\;\lim\limits_{x\to1}\dfrac{x^3+x-2}{x^3-x^2-x+1}\;$ but we can calculate the following two limits :

$\lim\limits_{x\to1^-}\dfrac{x^3\!+\!x\!-\!2}{x^3\!-\!x^2\!-\!x\!+\!1}=-\infty\;$ and $\;\lim\limits_{x\to1^+}\dfrac{x^3\!+\!x\!-\!2}{x^3\!-\!x^2\!-\!x\!+\!1}=+\infty\,.$

Now, we will calculate the first limit.

$\begin{align}\lim\limits_{x\to1^-}\dfrac{x^3+x-2}{x^3-x^2-x+1}&=\lim\limits_{x\to1^-}\dfrac{(x-1)(x^2+x+2)}{(x+1)(x-1)^2}=\\[3pt]&=\lim\limits_{x\to1^-}\dfrac{x^2+x+2}{(x+1)(x-1)}=\dfrac4{0^-}=-\infty\,.\end{align}$

Analogously, we can calculate the second limit.

Here is the graph of the function $\,f(x)=\dfrac{x^3+x-2}{x^3-x^2-x+1}\,.$

Evaluating $\lim_{x\to 1}\frac{x^3+x-2}{x^3-x^2-x+1}$

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