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I can't find this for some reason. I know I asked about 6 of these before, and I was able to finish my homework but now I went back to review and I can't do a single one of these problems on my own. Even the ones I did figure out on my own. I spent probably a total of 14 hours on the homework, not sure what is wrong with me but I don't know how to factor or do basic algebra.

Anyways I need to find $$\lim_{x\to -2}\frac{x+2}{x^3+8}.$$ I have spent at least an hour on it and I can't figure it out.

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HINT: $x^3+8 = (x+2)(x^2-2x+4)$. –  Arturo Magidin Aug 31 '11 at 21:15
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$(x+y)(x^2-xy+y^2)=x^3+y^3$... –  J. M. Aug 31 '11 at 21:15
    
Man I can't believe how bad I am at factoring. I tried that four an hour and I couldn't figure out how to force it to work. How do I look at that and know what is how to factor it? I assumed that the x squared would throw it off and mess it up. I am incredibly bad at factoring,starting to think math isn't for me. Also how do I know when to use a conjugat or not? I tried that and it didn't help. –  user138246 Aug 31 '11 at 21:16
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Are you allowed to use l'Hopital's? Are you familiar with the division algorithm for polynomials? –  Qiaochu Yuan Aug 31 '11 at 21:16
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@Jordan: For any polynomial $p(x)$, if $p(a)=0$, then $(x-a)$ divides $p(x)$. You can find the factorization by long division or by synthetic division. Here, since $x^3+8$ evaluates to $0$ at $-2$, then you must be able to factor $x+2$ out out $x^3+8$, and either of the two division methods will give you that the other factor is $x^2-2x+4$. –  Arturo Magidin Aug 31 '11 at 21:18

5 Answers 5

up vote 5 down vote accepted

A clever substitution might even gift you the answer.

It is usually easy to take the limit of a function about some special points, like $0$, $1$, $\pm \infty$ etc, since it is easier to get a feel for functions around such points. Also, most of the elementary limits like $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{e^x-1}{x} = 1$ already have the arguments approaching $0$ anyway, so it may be convenient if a new complicated limit looks like this as well.

Finally, as in any calculation, you should be especially careful while dealing with negative quantities. In your question, for instance, the negative sign in $-2$ could lead to some confusion while factoring, unless you are systematic in your long division or synthetic division.

In the light of the above points, I recommend the substitution strategy for evaluating limits. That is, if a given limit has $x \to a$, make the substitution $y = x-a$ and let $y \to 0$. Plug in $x=y+a$ in the function, and see if the resulting expression is simpler. As a rule of thumb, I would expect this trick would be powerful when you are dealing with polynomial or rational functions, or even exponential function $e^x$. On the other hand, if your limit involves trigonometric functions, then I would guess that the resulting expression would become more complicated, basically bacause the sum formulas of trigonometric functions involve multiple terms.

Let's see what this does in your example. Let $y = x+2$, so that $$x^3+8 = (y-2)^3+8 = y^3 - 8 - 3\cdot y \cdot 2(y-2) +8 = y^3 - 6y^2 + 12 y .$$ Hence, the given limit becomes: $$ \lim_{y \to 0} \frac{y}{y^3 - 6y^2 + 12y} = \lim_{y \to 0} \frac{1}{y^2 - 6y + 12} = \frac{1}{12} . $$ Notice that the evaluation of the limit in the final step is rather obvious.

Of course, I must warn you that you cannot blindly apply this (or, rather any) trick in all cases. For instance, given the limit $$ \lim_{x \to \pi/2} \frac{\exp(\cos x) - 1}{\cos x}, $$ rewriting it as $$ \lim_{y \to 0} \frac{\exp(\cos (y + \pi/2)) - 1}{\cos (y+\pi/2)} $$ seems hardly useful. (Of course, substitution is the way to go in this example, just not this one.)

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While there is nothing inherently wrong with this answer, given the obvious weakness in algebra by the OP, I suspect it is bound to be the source of far more confusion and despair than help, at least as far as he is concerned... –  Arturo Magidin Sep 1 '11 at 0:00
    
@Arturo I actually thought this one would be the most useful since the substitution is fairly mechanical. Are you afraid that the OP cannot correctly simplify an expression like $2(x-2)^3 - 3(x-2)^2 - 10$? In that case, factoring a polynomial (even given one factor) seems like a taller order... Just my 2 cents :) –  Srivatsan Sep 1 '11 at 0:04

From the comments, it appears that it would be useful to give a fairly elementary and pedantic discussion of using the factor theorem (from precalculus or college algebra, in the U.S.).

Factor Theorem: If $r$ is a zero of a polynomial $P(x)$ (i.e. a solution to $P(x) = 0$ is $x=r$), then $x-r$ is a factor of $P(x)$.

Example 1: Factor $x^3 - 8$.

First, solve $x^3 - 8 = 0$ to find a value for $r$. Solving gives $x^3 = 8$, or $x = 2$. Therefore, we can use $r = 2$, which tells us that $x - 2$ is a factor of $x^3 - 8$. Now use long division (or synthetic division) to determine the quotient when $x - 2$ is divided into $x^3 - 8$. The quotient will be $x^2 + 2x + 4$. Therefore, we can factor $x^3 - 8$ as $(x-2)(x^2+2x+4).$

Example 2: Factor $x^3 + 8$.

First, solve $x^3 + 8 = 0$ to find a value for $r$. Solving gives $x^3 = -8$, or $x = -2$. Therefore, we can use $r = -2$, which tells us that $x - (-2)$, or $x+2$ is a factor of $x^3 + 8$. Now use long division (or synthetic division) to determine the quotient when $x + 2$ is divided into $x^3 + 8$. The quotient will be $x^2 - 2x + 4.$ Therefore, we can factor $x^3 + 8$ as $(x+2)(x^2-2x+4).$

Example 3: Factor $x^5 - 100,000$.

First, solve $x^5 - 100,000 = 0$ to find a value for $r$. Solving gives $x^5 = 100,000$, or $x = 10$. Therefore, we can use $r = 10$, which tells us that $x - 10$ is a factor of $x^5 - 100,000$. Now use long division (or synthetic division) to determine the quotient when $x - 10$ is divided into $x^5 + 100,000$. The quotient will be $x^4 + 10x^3 + 100x^2 + 1000x + 10,000$. Therefore, we can factor $x^5 - 100,000$ as $(x-10)(x^4 + 10x^3 + 100x^2 + 1000x + 10,000).$

Example 4: Factor $x^5 + x^4 + x^3 + x^2 + x + 1$.

First, solve $x^5 + x^4 + x^3 + x^2 + x + 1 = 0$ to find a value for $r$. In this case, standard methods don't work (unless you know about cyclotomic equations). However, by using the rational root test (google it), you'll get $x = 1$ and $x = -1$ as possible solutions. You'll find by direct substitution that $x = -1$ is a solution to $x^5 + x^4 + x^3 + x^2 + x + 1 = 0$. Therefore, we can use $r = -1$, which tells us that $x - (-1)$, or $x + 1$ is a factor of $x^5 + x^4 + x^3 + x^2 + x + 1$. Now use long division (or synthetic division) to determine the quotient when $x + 1$ is divided into $x^5 + x^4 + x^3 + x^2 + x + 1.$ The quotient will be $x^4 + x^2 + 1$. Therefore, we can factor $x^5 + x^4 + x^3 + x^2 + x + 1$ as $(x+1)(x^4 + x^2 + 1).$

Here's the really nice thing about your situation. There's almost no work in trying to find a solution to $P(x)= 0$. For example, in your specific limit problem you know that $x^3 + 8 = 0$ when $x=-2$ (because you presumably already tried to find the limit by plugging in $x=-2$). So you already know a value that can be used for $r$ in the factor theorem. In other words, something "tricky" to solve like Example 4 won't come up.

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For $x=-2$, $$\begin{eqnarray*} x+2 &=&0 \\ x^{3}+8 &=&0, \end{eqnarray*}$$ which means $x=-2$ is a root of both equations. Hence $x^{3}+8$ may be factored$^1$ as $$x^{3}+8=(x-(-2))Q(x)=(x+2)Q(x).\qquad(*)$$

To compute $Q(x)$

  • you may perform the long division $(x^{3}+8):(x+2)$ (or apply the so called Riffini's rule or sinthetic division) and find $$Q(x)=x^{2}-2x+4.$$

  • Alternatively you write $x^{3}+8$ as $$x^{3}+8=(x+2)(ax^{2}+bx+c),$$ (you could have made $a=1$, because the coefficient of $x^3$ is $1$) expand the RHS $$(x+2)(ax^{2}+bx+c)=ax^{3}+bx^{2}+cx+2ax^{2}+2bx+2c,$$ group the terms of the same degree $$(x+2)(ax^{2}+bx+c)=ax^{3}+\left( b+2a\right) x^{2}+\left( c+2b\right) x+2c$$ and equate to $x^{3}+8$ $$ax^{3}+\left( b+2a\right) x^{2}+\left( c+2b\right) x+2c=x^{3}+8.$$ The two sides are the same polynomial if and only if their coefficients are equal $$\begin{eqnarray*} a &=&1 \\ b+2a &=&0 \\ c+2b &=&0 \\ 2c &=&8, \end{eqnarray*}$$ which means $a=1,b=-2,c=4$.

So $$x^{3}+8=(x+2)(x^{2}-2x+4).\qquad(**)$$

Thus, for $x\neq -2$, we have $$ \frac{x+2}{x^{3}+8}=\frac{x+2}{\left( x+2\right) \left( x^{2}-2x+4\right) }= \frac{1}{x^{2}-2x+4},\qquad(***)$$ i.e $\frac{x+2}{x^{3}+8}$ is equal to $\frac{1}{x^{2}-2x+4}$, except for $x=-2 $. Now you can safely compute the limit

$$\lim_{x\rightarrow -2}\frac{x+2}{x^{3}+8}=\lim_{x\rightarrow -2}\frac{1}{ x^{2}-2x+4}=\dfrac{1}{\displaystyle\lim_{x\rightarrow -2} x^{2}-2x+4}=\dfrac{1}{(-2)^{2}-2(-2)+4}=\dfrac{1}{12}.$$

--

$^1$ If a polynomial of degree $n$,

$$P(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\ldots +a_{n-1}x+a_{n}\quad (a_{0}\neq 0),$$

has as roots $n$ different numbers $\alpha _{1},\alpha _{2},\ldots ,\alpha _{n}$ , then it may be factored as

$$P(x)=a_{0}(x-\alpha _{1})(x-\alpha _{2})\cdots (x-\alpha _{n}).$$

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HINT $\ $ Consider the limit of a rational function $\rm\:f(x)/g(x)\:$ as $\rm\:x\to c\:.\:$ If $\rm\:f(c)=0=g(c)\:$ then, by the Factor Theorem, both $\rm\:f,g\:$ are divisible by $\rm\:x-c\:.\:$ Keep cancelling $\rm\:x-c\:$ from both $\rm\:f,g\:$ till this is no longer possible. Then the limit will no longer be of indeterminate form $\:0/0$.

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I don't really understand anything you just said. What is a limit point? What is x=c? I don't know what (x-c)^N is. I do not know what division algorithm or euclidan algorithm is either. –  user138246 Aug 31 '11 at 21:37
    
@Jordan Hopefully you can understand it now. –  Bill Dubuque Aug 31 '11 at 21:48
    
Thanks I get it now I think. –  user138246 Aug 31 '11 at 22:26

Are you allowed to use l'Hôpital's rule? If you plug $x=-2$ into your formula, you have $0/0$ which is an indeterminate form that l'Hôpital's rule can handle.

Taking derivatives of the numerator and denominator, we get $$ \frac{(x+2)'}{(x^3+8)'} =\frac{1}{3x^2}$$ and now plugging in $x=-2$ we have $1/12$ as the answer.

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I don't know what that is. –  user138246 Aug 31 '11 at 21:21
    
It's a simple rule to help evaluate certain types of limits. If you haven't seen it yet, you will soon (in Calculus I, typically). Anyway, use the factoring method instead since that's obviously what your teacher is intending you to do here. –  Fixee Aug 31 '11 at 21:22
    
@Jor For rational functions, L'Hospitals rule is essentially equivalent to cancelling common factors - as explained in my answer. –  Bill Dubuque Aug 31 '11 at 21:56

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