Recognizing the proper polynomial factorization to solve an indeterminate limit

Question

I had to solve the $\lim_{x \to 3} \frac{x^3-3x^2-x+3}{x^2-x-6}$ that is indeed an indeterminate form ($\frac{0}{0}$).

The approach I adopted was to factor the polinomials so that I can deviate from the indeterminate form.

$$ \lim_{x \to 3} \frac{x^3-3x^2-x+3}{x^2-x-6} = \lim_{x \to 3} \frac{(x-3)(x^2-1)}{(x-3)(x+2)} = \lim_{x \to 3} \frac{(x^2-1)}{(x+2)} = \frac{8}{5} $$

Unfortunately, my first approach was to factor the denominator into $(x-2)(x+3)$ that didn't lead to the same simple solution.

Hence my question. Is there any rule, hint or suggestion I can keep in mind in cases like this one to properly determine the most appropriate factorization to simplify the fraction and the calculus?

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EDIT: I just noticed, reading my own question, that the first attempt was not as simple because the factorization itself was wrong. Please ignore the question.

Answer 1

Note that $(x-2)(x+3) = x^{2} + x - 6 \ne x^{2} - x - 6 = (x + 2 ) (x - 3)$. Polynomials of degree $1$ are irreducible, so factorization is unique here.

Answer 2

The factorization of a polynomial is essentially unique. Your first attemp is wrong in signs.

Answer 3

There is only one way to factorise $x^2-x-6$. $(x-2)(x+3)$ did not work because it is wrong.

Answer 4

Your first approach apparently had a sign error, because $\,(x-3)(x+2) \ne (x+3)(x-2),\,$ since $\,x=3\,$ is a root of the LHS but not the RHS. The rule to keep in mind is this. If you have a quotient of polynomials $\,f/g\,$ that is of indeterminate form $\,0/0\,$ at $\, x = a\,$ then, by applying the the Factor Theorem, we deduce that $\,x-a\,$ is a common factor of both $f$ and $g$, so we can cancel the common factor $\,x-a\,$ then try evaluating again. If that is still of form $\,0/0\,$ then we can cancel $\,x-a\,$ again. Iterating this process, we eventually reach the point where $f$ and $g$ are not both $0$ at $x= a$, so the limit is computable by evaluation (specially handling the case $\,g(0) = 0).$ One could also reduce $\,f/g\,$ to lowest terms by cancelling $\,\gcd(f,g)\,$ but that may be more work than needed, since to reach a determinate form requires cancelling only $\,x-a\,$ factors, not all common factors.