# Limit on a five term polynomial

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: $$\lim_{x \to\ 1}\frac{{x^4+3x^3-13x^2-27x+36}}{x^2+3x-4}$$

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

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

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!

• There was an arithmetical error in the long division. If done correctly we get quotient $x^2-9$ and remainder $0$. – André Nicolas Jan 31 '16 at 4:30

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.
Ok, so I figured out my error. I was allowing the term $$13x^2$$ 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,x^4+3x^3-$$13x^2$$-27x+36becomes $$x^4+3x^3-9x^2-4x^2-27x+36$$Next, I group the terms that have common factors. Thus the previous term becomes $$x^2(x^2+3x-4)-9(x^2+3x-4)$$Since the term $$(x^2+3x-4)$$ is identical to the denominator, the only terms left are $$x^2-9$$Hence, \lim_{x \to\ 1}$$x^2-9$$=-8