How to find the following limit algebraically?

Question

I've been trying to answer this for a while and I know it's a simple question relative to most questions that are posted here.

$$ \lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)} $$

If we substitute -2 for $ x $ we get $ 0/0 $, an indeterminate form. I figured that the denominator can be rewritten as $ (x^2-4)(x+1) $. And then I tried to factor something in the numerator but couldn't see anything interesting. How do I find the limit algebraically?

I know that the answer is supposed to be 1, but I don't know how they got there.

Thanks in advance!

Answer 1

General tip (update) : When you can see that the denominator is equal to zero for a value $x=a$ which is the $x\to a$ of the limit, then you should try factoring on both the numerator and the denominator the factor $(x-a)$, such as you can get rid of the $\frac{0}{0}$ issue. This particular example though can also be done by following a row factorization.

Factor the expression at the numerator by taking out $x^2$ and then forming a quadratic factorization inside, as :

$$x^4+5x^3+6x^2 = x^2(x^2+5x+6) =x^2(x^2+2x+3x+6) = x^2(x+2)(x+3)$$

Then, the given limit is :

$$\lim_{x\rightarrow -2} \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)} = \lim_{x\to -2} \frac{x^2(x^2 + 5x + 6)}{(x^2-4)(x+1)} = \lim_{x \to -2} \frac{x^2(x+3)(x+2)}{(x-2)(x+2)(x+1)} $$

$$=$$

$$\lim_{x\to -2} \frac{x^2(x+3)}{(x-2)(x+1)} = 1$$

Answer 2

HINT: $x^4+5x^3+6x^2 = x^2(x^2+5x+6) =x^2(x^2+2x+3x+6) = x^2(x+2)(x+3)$

Answer 3

Write $$\frac{x^4+5x^3+6x^2}{(x^2-4)(x+1)}=\frac{x^2(x+3)(x+2)}{(x-2)(x+2)(x+1)}.$$

Answer 4

Notice that: $$x^4+5x^3+6x^2 = x^2(x^2+5x+6) = x^2(x+2)(x+3)$$ and $$x^2(x+1)-4(x+1) = (x^2-4)(x+1) = (x+2)(x-2)(x+1).$$

Then, you have:

$$\lim_{x\rightarrow -2}\: \frac{x^2(x+2)(x+3)}{(x+2)(x-2)(x+1)} = \\ = \lim_{x\rightarrow -2}\: \frac{x^2(x+3)}{(x-2)(x+1)} = \\ = \frac{(-2)^2(-2+3)}{(-2-2)(-2+1)} = \frac{4}{4} = 1.$$

Answer 5

Let $x+2=h\iff x=h-2$

$$\lim_{h\to0}\dfrac{(h-2)^4+5(h-2)^3+6(h-2)^2}{(h-2)^3+(h-2)^2-4(h-2)-4}$$

$$=\lim_{h\to0}(h-2)^2\cdot\lim_{h\to0}\dfrac{(h-2)^2+5(h-2)+6}{h^3-5h^2+h(12-4-4)}$$

$$=(0-2)^2\cdot\lim_{h\to0}\dfrac{h(1+h)}{h(h^2-5h+4)}=?$$

Answer 6

What, nobody used L'Hôpital's rule yet? Couldn't resist:

$$\lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)} = \lim_{x\rightarrow -2}\: \frac{4x^3+15x^2+12x}{3x^2+2x-4} = \frac{4}{4}$$

Answer 7

The fact the numerator becomes zero after plugging in $x = -2$ means that $(x + 2)$ is a factor of the numerator; that is the "interesting" thing you can factor out. You can factor it out of the denominator for the same reason.

Doing so gives

$$ \lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)} = \lim_{x\rightarrow -2}\: \frac{(x+2)(x^3 + 3x^2)}{(x+2)(x^2 - x - 2)} = \lim_{x\rightarrow -2}\: \frac{x^3 + 3x^2}{x^2 - x - 2} $$

and this limit can be found by plugging in $x = -2$.