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it has been two years since I have taken or used calculus, and I am having some trouble with factoring a polynomial in order to take a limit on it. I have searched for previous similar questions here, but I have been unable to find anything helpful. Here is my problem:

I have: \begin{equation} \lim_{x \to\ 1}\frac{{x^4+3x^3-13x^2-27x+36}}{x^2+3x-4} \end{equation}

So, how do I approach this? When I try long division, I end up with a remainder, e.g.:

\begin{equation} _{x \to\ 4}\frac{{x^4+3x^3-13x^2-27x+36}}{x^2+3x-4} = (x+3)(x+3)+\frac{{24x+72}}{(x+1)(x-4)} \end{equation}

Plugging in the limit to what I came up with via long division just yields an undefined result, so I'm obviously doing something wrong.

If anyone can help, I would really appreciate it!

Thanks!

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  • $\begingroup$ There was an arithmetical error in the long division. If done correctly we get quotient $x^2-9$ and remainder $0$. $\endgroup$ – André Nicolas Jan 31 '16 at 4:30
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Since the numerator is non-zero when $x = 4$, the limit does not exist. We can write the rational function as

$$R(x) \cdot \frac{1}{x - 4}$$

where $R(4) \ne 0$ (and $R$ is continuous on a neighborhood of $4$). But since

$$\lim_{x \to 4} \frac{1}{x - 4}$$ does not exist, your limit does not exist.

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  • $\begingroup$ Thanks a lot! Unfortunately, I realized that I wrote the wrong limit in my original post, so that would explain why you came up with DNE for the limit. Thanks again though! $\endgroup$ – Sage Hopkins Jan 31 '16 at 3:49
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Ok, so I figured out my error. I was allowing the term \begin{equation}13x^2\end{equation} from the numerator to mess me up. Factoring the numerator needs to be approached by first turning the prime number, 13, into non-prime numbers. Thus,\begin{equation}x^4+3x^3-13x^2-27x+36\end{equation}becomes \begin{equation}x^4+3x^3-9x^2-4x^2-27x+36\end{equation}Next, I group the terms that have common factors. Thus the previous term becomes \begin{equation}x^2(x^2+3x-4)-9(x^2+3x-4)\end{equation}Since the term \begin{equation} (x^2+3x-4)\end{equation} is identical to the denominator, the only terms left are \begin{equation}x^2-9\end{equation}Hence, \begin{equation}\lim_{x \to\ 1}x^2-9=-8\end{equation}

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