Using limits to investigate problems

Question

$\cot(π/2)$ is said to be $0$ rather than undefined, even though it equals $1/\tan(π/2)$ and $\tan (π/2)$ is undefined.

Some people also say that cotangent is "base divided by perpendicular": for $\theta=π/2,$ the base is $0,$ so cotangent equals $0.$ Using the concept of limits, other people argue that $\theta$ is not equal to $π/2$ but very close to it: $\cot(\theta)$ is approaching $0$ from both sides, so $\cot(\theta)$ is equal to $0$ at $\theta=π/2.$

Okay I am convinced by this second argument. But now consider the exercise $$\frac{a}{b+c} = \frac{b}{c+a} = \frac{c}{b+a} = \frac{1}{k}.$$ Writing separate equations like $a =\frac{b+c}{k}$ then addding them, we get $(a + b + c)= \frac{2(a + b + c)}{k}$ and finally $k=2,$ though only if $a+b+c\not=0.$ If we use the concept of limits, we see that as $a+b+c$ approaches $0,$ $k$ is heading towards $2.$ However, if $a=1, b=1, c=-2,$ then $k=-1$ is another answer. What's going on? Why do we sometimes use the concept of limits but sometimes not, for similar kind of problems?

Answer 1

$\cot(π/2)$ is said to be $0$ rather than undefined, even though it equals $1/\tan(π/2)$ and $\tan (π/2)$ is undefined.

We have the standard definition $$\cot\theta:=\frac{\cos\theta}{\sin\theta}\tag1$$ and the trigonometric identity (an identity is an equation that holds whenever its both sides are defined) $$\cot\theta\equiv\frac1{\tan\theta}.\tag2$$

Definition $(1)$ gives $$\cot\frac{\pi}2=\frac01=0,$$ whereas for $\theta=\dfrac{\pi}2,$ identity $(2)$ does not apply.

(To be clear: cotangent is not generally defined as the reciprocal of tangent.)

Answer 2

There are many ways to define a function such as cotangent. Some of them don’t need limits at all.

If someone uses a limit in their definition of cotangent, it’s just in order to define a function with the same values as everyone else’s definition of the cotangent. For example, here are three definitions that don’t require limits:

$$ \cot(x) = \frac{\cos(x)}{\sin(x)},$$ $$ \cot(x) = \tan\left(\frac\pi2 - x\right),$$ $$ \cot(x) = \begin{cases} \frac1{\tan(x)} & \text{$x$ not an integer multiple of $\frac\pi2$,}\\ 0 & \text{$x$ an integer multiple of $\pi$.} \end{cases} $$

It is also possible to define $\cot(x)$ using a power series, which means every value of the function is a limit (of the partial sum of a series as the number of terms goes to infinity).

No matter how you define the function, however, a definition of a function such as $\cot$ is a definition, which means we get to do whatever we need to do in order to make the definition be useful. In the case of the cotangent, part of being useful is agreeing with conventional definitions, but if you’re defining a new function you don’t have even that constraint.

On the other hand, solving an equation typically looks for all solutions. For your equation in $a,b,c,k$ you can consider two cases, $a+b+c=1$ and $a+b+c\neq 1$, and find the complete solution set. That set does not define $k$ as a function of $a,b,c$. That’s just how it is. You can still define a function using part of that solution set if you have a need for such a function. You don’t need limits to do so, but if you want your function to be continuous then limits are technically involved everywhere in the definition.

Answer 3

You've got to use the right tool for each job. You first need to know precisely what it is you want to do.

To compute a limit, you need a function and a topology for its domain and codomain (e.g. a real function). In your first example, we have a function $1/\tan$ that is not defined, a priori, on $\pi/2$. But we could "extend" this function, adding a value for this argument. If you want your extended function to still be continuous, then the value you have to fill in has to be the limit of your original function at that point.

In your second example, we are asking what is the set of solutions of a certain equation, in particular the set of values of $k$ that satisfy the equation for a given value of $a + b + c$. This is not a real-valued function, because for a given value of $a + b + c$ it gives a set of values of $k$: for nonzero values of the former, it just happens that this set contains a single element, 2, while when the input is zero the output is the set $\mathbb{R}\setminus \{0\}$. So it doesn't make much sense to ask about limits here, because we're working with sets of real numbers, not real numbers. (Unless you define a topology for this.)

Of course, the limit of the function that is 2 everywhere except at 0 is 2, but do you see that this is not the complete picture of the problem you were solving? You are choosing a value out of each solution set and forming a function with that. Its limit will also likely be a solution (depending on the continuity of operations used in your equation etc.), but it will not account for all solutions.