# How to find the following limit algebraically?

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.

• Divide top and bottom by the highest power. – uniquesolution Jan 26 '18 at 10:05
• @uniquesolution That only works for limit as $x\to \pm \infty$. Or, at least, that's when your suggested approach is the standard, simple approach. – Arthur Jan 26 '18 at 10:08
• @Arthur Right on point, +1 ! – Rebellos Jan 26 '18 at 10:11

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 :

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

$$=$$

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

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 $x = -2$.

• +1 - It is surprising that you are the only answerer so far to bring up this simple and obvious fact: The very issue here - that the numerator and denominator are both $0$ at $x = -2$ - tells you the common factor you can divide from both to find the limit. You don't need to fully factor the polynomials. – Paul Sinclair Jan 26 '18 at 17:39
• @PaulSinclair (+1) I completely agree. It would be kind of fun to come up with an example involving "ugly" polynomials in the numerator and denominator for $\lim_{x\to a}\frac{P(x)}{Q(x)}$ where you couldn't factor the polynomials nicely and where you were more or less forced to use synthetic division and the factor theorem to factor out the factors $x-a$ appropriately and then evaluate as Hurykl did. – Daniel W. Farlow Jan 26 '18 at 20:19
• @DanielW.Farlow : Such a pair of "ugly" polynomials would fall to the slightly fancier method of using Hurkyl's method. The polynomial GCD of the given numerator and denominator is $\gcd(x^4+5 x^3+6 x^2, (x^2-4)(x+1)) = x+2$. – Eric Towers Jan 27 '18 at 7:25
• @EricTowers Maybe I'm misunderstanding your comment or perhaps I did't clearly express what I had in mind in my comment. Basically what I meant to say was trying to force someone to only use the "slightly fancier method," where it might be rather difficult to completely factor the rest of the polynomial. I'm not sure what your comment really addresses in my own comment unless I am simply overlooking something. – Daniel W. Farlow Jan 27 '18 at 7:43
• @DanielW.Farlow : You can't force synthetic division and the factor theorem because you can't make a problem so ugly that it fails to fall to the "simpler" process of taking the polynomial GCD. Polynomial GCDs are just too darned easy. – Eric Towers Jan 27 '18 at 7:46

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

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

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

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

• Welcome to MSE. What's algebraic about your approach? – José Carlos Santos Jan 26 '18 at 12:05
• Sorry, my maths education is apparently a bit different from what it's like in English-speaking countries, so I may be unfamiliar with this particular use of the term "algebraic". You can ignore this answer; like I said, I just couldn't resist trying to solve it this way, even if it might not have been particularly useful as more than equation-writing practice. – Ove Jan 26 '18 at 12:34
• @Ove, your desire is understandable, but it is implicit in the question that the exercise is leading up to applied calculus. Applying a result like L'Hopital defeats the exercise (and doesn't help the student). – user1717828 Jan 26 '18 at 17:27
• Well, as I mentioned, the word "algebraically" meant nothing to me in this context, in part because limits aren't pure algebra anyway. I've done a lot of calculus without knowing that in the English-speaking world, people apparently use this word like this. Do you want me to delete this answer? – Ove Jan 27 '18 at 0:03
• @user1717828 technically this is algebra. asker should clarify what rules they limit themselves to first. – The Great Duck Jan 27 '18 at 4:36

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

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

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

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

• @Downvoter, Any apparent mistake? – lab bhattacharjee Jan 26 '18 at 10:19
• My be your way is uncommon – E.H.E Jan 26 '18 at 10:41
• @E.H.E that is more of a reason to upvote. – The Great Duck Jan 27 '18 at 4:37

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:

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