Math question - limits

I’ve applied to return to university and am meantime brushing up on my calculus. In an old textbook, I found this problem but no explanation of how to resolve it. I’m sure it’s quite obvious and easy and I don’t have problems resolving other polynomials, but this one escapes me:

x -> -a

x[sup]3[/sup] + a[sup]3[/sup]

x + a

The given answer is3a[sup]2[/sup]but I can’t grasp the intervening steps. Thanks.

If I remember my calc classes right, the trick for those is usually to factor the top and pull out a term equal to the bottom, cancel that factor (to remove the zero in the denominator), then apply the limit.

Of course, I can’t remember how to factor cubes. :slight_smile: A quick google found this, which makes the steps:

lim x -> -a: (x^3 + a^3) / (x + a)
lim x -> -a: (x + a ) * ( x^2 - ax + a^2) / (x + a)
lim x -> -a: x^2 - ax + a^2

At this point, you’ve removed the divide by zero, so just replace x with -a:
(-a)^2 - a * (-a) + a^2
a^2 + a^2 + a^2

sciguy got it, but there’s another way to explain it using a method that is more generally applicable: l’Hopital’s rule. Essentially, you take the derivative of the top and bottom, separately, and then restore the fraction. Like so:

lim ( 3x[sup]2[/sup])/(1)

Which is easily evaluated at 3a[sup]2[/sup].

I tried doing the long division procedure but got stuck until sciguys’s post reminded me to put in dummy values, changing this:

(x+a) | x^3 + a^3
to this:

(x+a) | x^3 + (0 x^2 a) + (0 x a^2) + a^3


x^2 - xa + a^2

…which made it easy, as I knew it had to be. As for l’Hopital, I guess that’s covered later in the book so I’ll flip ahead. Thanks, guys.

It should be noted that l’Hopital’s rule only works in a few specific cases. It only works for indeterminate forms, like 0/0, infinity/infinity, and 0*infinity. Of course, anything that isn’t an indeterminate form, there are probably easier methods to use instead.

Moreover, is it strictly kosher to use differentiation to solve a limit? After all, differentiation is based on the theory of limits, right?

Chronos is, of course, correct. It’s for indeterminate forms. As to Spectre’s question, it works because when you take the derivatives separately, each one is not an indeterminate form, so the limit works just fine. Think of it as breaking a very complex limit problem into two simpler limit problems.

Another way to think of it (and this way comes in handy from time to time):

This limit has the form of the definition of the derivative of x^3 evaluated at -a. The deriviative of x^3 is 3x^2. Plug in -a and you get 3a^2.

It’s not unusual for calculus books to have limit questions that are designed to be solved by recognizing the limit as the definition of the derivative of a specific function.

For example:

lim (as x goes to a) of [sin(x)-sin(a)] / [x - a]

If you recognize that as the derivative of sin(x) evaluated at x=a, it’s easy to see the limit is cos(a).

It is if you’re correctly applying L’Hopital’s rule.

However, in terms of logical dependence, L’Hopital’s rule involves derivatives, which are defined in terms of limits, which is why most calculus books introduce limits first (and evaluating them by factoring, as in sciguy’s response), then derivatives, and only later do they introduce L’Hopital’s Rule, which makes some people think, “Gee, it would have been handy to have that back when we were doing all those limits!”

Just because the definition of a derivative involves taking a limit doesn’t mean this is circular reasoning or anything. You’re just evaluating a single limit. l’Hôpital’s rule is perfectly fine.

It may not be circular, but surely violates the logical dependency for someone learning. What bothers me about all these posts is that limits have a meaning and they all treat it as a formal exercise. (In this, they are probably no worse than the authors of the book.) Much better is to look at a specific example, say the limit as x → -2 of (x^3 + 8)/(x + 2). Then evaluate the quotient when x = -3, x = -2.1, x = -2.01 until you get the idea that it is tending to 12. The three values above are 19, 12.61, 12.06. Then do it for a = 3 and then try to see the general pattern. After that you can try long division and reduce the fraction.

I will tell you this: if a beginning calc student had used l’Hospital’s rule to solve this problem, he’d have gotten no credit. This would almost invariably be an AP student who believes he knows calc and actually doesn’t know beans.

Just to check: you mean, “a student learning how to evaluate limits for the first time does not have access to l’Hôpital’s rule”, or are you making a claim about the pedagogy?

I parsed Spectre’s question as “Is that a valid mathematical technique (at all)?” rather than “Is that a tool a student has access to at this point?” If that’s not what he meant, my apologies.

The first. Since they learn limits (or, mostly, don’t) before calculus, they don’t have access to l’Hospital’s rule. Moreover, if they do know l’Hospital (from AP or otherwise) and used it, they still wouldn’t learn anything about limits.

So, yes, my point is essentially pedagogical. But then I am–or was–a pedagogue.

Here is how I understand limits. The limit of f(x) and x → a is the ordinary part of f(a+h) when h is infinitesimal, assuming that this ordinary part doesn’t depend on which infinitesimal. In the problem at hand, you get 3a^2 - 3ah and since 3ah is infinitesimal when h is (and a is finite), the ordinary part is 3a^2. Of course, the whole computation depends on doing that division.

Whoa, hey, I just asked a math question. Don’t let’s get divisive.

The way I like to think of L’Hopital is that you have x/y and x and y are both approaching zero, which gives you an indeterminate form. The question is "Which one approaches zero FASTER? It might not seem like it matters, because they’re both getting infinitesimally close to zero eventually. But if x is getting to zero faster than y, your limit will be zero. If y is getting there faster, it will be infinity or some finite value. If they get to zero at the same rate, the limit will be one. (Trivial example, limit as x->0 of x/x is obviously 1)

This isn’t a rigorous proof, of course, but I think it’s nice to understand things intuitively too.

Yeah, that’s the basic intuitive idea.

A slightly more formal (yet still not rigorous) explanation of the intuitive idea:

You have the limit as x approaches a of the quotient f(x)/g(x), with f(a)=g(a)=0.

The idea behind L’Hospital’s rule in this case is to use a tangent line to both f and g to approximate the functions.

The tangent line to f(x) at x=a is f ’ (a)(x - a) (and for g(x) it is g ’ (a)(x - a) ).

So rather than look at the limit as x approaches a of f(x)/g(x), approximate both f and g with their tangent lines:

The limit as x approaches a of

f ’ (a)(x - a)

g’ (a)(x - a)
which simplifies to f ’ (a) / g ’ (a).