So I was trying to evaluate the following limit:
$$f(x) = \lim \limits_{x\to 3} \frac{x^4 - 3x^3}{x^4-81}$$
To do so, you would need to factor the denominator to $(x^2+9)(x+3)(x-3)$ and cancel out $(x-3)$. If I wouldn't have seen it being factored this way, it would've been pretty difficult for me to factor the denominator in 3 terms. Is there a simple way/method to factor cases like that, that would work in most cases?
Thanks in advance!
On
Hopefully you know that $a^2-b^2=(a+b)(a-b)$
We can use this on the initial equation to get $$x^4-81=(x^2+9)(x^2-9)$$
We can then use the same technique on the second bracket to get $$x^2-9=(x+3)(x-3)$$
This gives us our final answer $$x^4-81=(x^2+9)(x+3)(x-3)$$
On
The first thing you try to evaluate the limit as $x\to 3$ is to boldly try and plug in $x=3$. You will notice that both numerator and denominator evaluate to $0$ there. This implies that performaing polynomial division by $x-3$ will succeed without remainder for both, whereafter you can cancel.
On
In this particular problem, as noted by others, it is probably simplest to factor using difference of two squares twice.
More generally, though, if you are taking the limit of a quotient of polynomials as $x\to c$ and get a $\frac00$ form, then since $c$ is a zero of the top and of the bottom, the Factor Theorem says that $x-c$ will be a factor of both the top and bottom. This factor can be taken out, perhaps with synthetic division, and cancelled. (You don't need to fully factor the numerator and denominator necessarily, just get them to the point of being able to cancel the $x-c$ factor.)
One of my favored trick when I see a limit for $x\to a$ is to make $x=y+a$ and work the limit for $y\to 0$.
So, in your case $$\frac{x^4 - 3x^3}{x^4-81}=\frac{27 y+27 y^2+9 y^3+y^4}{108 y+54 y^2+12 y^3+y^4}\to \frac {27}{108}=\frac 14$$