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In Order to find the limit of a fraction with polynomial in numerator and denominator we can simplify the fraction by finding the common factor in numerator and denominator, canceling them out and solving the limit.

My question is, in a polynomial with degree $n$, which $x$ should I factor out? For example in the fraction

$$\lim_{x \to \alpha} \frac{5x^2 + x + 3}{4x^3 + x^2 + 1}$$

I have the following options:

  1. Choosing one common factor for the whole fraction

    1.1 Choosing $x$ with the highest degree ($x^3$) in both numerator and denominator and factoring out

    1.2 Choosing $x$ with the lowest degree ($x$) in both numerator and denominator and factoring out

  2. Choosing the common factor in numerator and denominator independently

    2.1. Choosing $x$ with the highest degree ($x^2$) in the numerator and $x$ with the highest degree ($x^3$) in the denominator and factoring out

    2.2. Choosing $x$ with the lowest degree ($x$) in the numerator and $x$ with the lowest degree ($x^2$) in the denominator and factoring out

Which case should I use? Does it depend on the approaching point $a$ of the limit? If yes, which case is used for which approaching point?

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The polynomial remainder theorem says that $a$ is a root of $f(x)$ iff $x-a$ is a factor of $f(x)$.

If we wanted to evaluate the limit $$\lim_{x\to a}\frac{p(x)}{q(x)}$$ we can typically do what we usually do and try to substitute $x=a$ (i.e. evaluate $\dfrac{p(a)}{q(a)}$).

If $p(a)=q(a)=0$, then factor out $x-a$ from both $p(x)$ and $q(x)$, cancel, and then substitute $x=a$.
If $p(a)\neq 0$ and $q(a)=0$, then the limit does not exist.
If $q(a)\neq 0$, then the limit is $\dfrac{p(a)}{q(a)}$.

If we wanted to evaluate the limit $$\lim_{x\to\infty}\frac{p(x)}{q(x)}$$ we would use the known limit $$\lim_{x\to\infty}\frac1x=0.$$

To use this, we would take the highest degree term in the denominator and divide each term by that term. Then, using properties of limits and the above limit, the entire limit can be evaluated. Here is an answer I have written to another question that will hopefully give you some insight.

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