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I am sort of lost when performing the limits of an equation.
For instance, let's take: $\lim\limits_{x\to-\infty} x^2-7x+12$
$(-\infty)^2-7(-\infty)+12$
$(\infty)+(\infty) \rightarrow$ Thus, it is indeterminate.
In this case, we cannot resort to L'hoptal's rule since we don't have indeterminate forms of $\frac{0}{0}$ or $\frac{\infty}{\infty}$
I thought of another method to solve this:
We factorise $x^2-7x+12$ into $(x)(x-7+\frac{12}{x})$
$(-\infty)(-\infty-7+\frac{12}{-\infty})$
$(-\infty)(-\infty+0)$
$\infty$
Could this method be a correct one, or was it coincidental that I got $\infty$ here?

What are other methods I can rely on other than L'hopital's rule in case I found myself in this situation? So far, I only know L'hopital's rule and taking the highest degree.
For instance, $\lim\limits_{x\to-\infty} x^2-7x+12$'s highest degree is $x^2$:
$\lim\limits_{x\to-\infty}x^2=(-\infty)^2=\infty$

Angelo
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Vocelia
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    $x^{2}(1-\frac 7 x+\frac {12} {x^{2}}) \to \infty$. – Kavi Rama Murthy Jun 02 '23 at 08:12
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    Notice that $$a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=a_nx^n\underbrace{\left(1+\frac{a_{n-1}}{a_{n}x}+...+\frac{a_1}{a_nx^{n-1}}+\frac{a_0}{a_nx^n}\right)}{\to 1\ if\ x\to \pm\infty }.$$ Therefore $$\lim{x\to \pm \infty }a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=\lim_{x\to \pm \infty }a_nx^n.$$ – Surb Jun 02 '23 at 08:13
  • Oh, I see! So, that method is indeed correct, but it is quite useless, as we can just simply take the highest degree, according to the proof you provided. – Vocelia Jun 02 '23 at 08:22

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