# Using limits to investigate problems

$$\cot(π/2)$$ 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?

• $\cot\frac{\pi}{2}=0$ because of how $\cot{\theta}$ is defined geometrically (not as the reciprocal of $\tan{\theta}$). Commented Aug 11 at 9:06
• Very simply speaking, the concept of limit is a topological notion that requires a minimal structure in the set in which you are working, that is, we at least want to have a concept of sequence of points and neighborhoods. For objects without such a context, the use of the limit does not make sense. In the case of trigonometric functions defined as ratios (such as cot and tan), as functions of a real variable, they are not defined at the points where the denominator is zero, but at such points we can still give a sensible definition that derives from their geometric construction . Commented Aug 11 at 9:18
• You're using MathJax incorrectly. Use it for the whole math expressions. Instead of writing a = $\frac{b+c}{k}$ write $a = \frac{b+c}{k}$. Also, avoid the use of unusual characters like π that won't render properly in some devices. Instead use MathJax and write it like \pi Commented Aug 11 at 11:25

$$\cot(π/2)$$ 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.)

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.

• Do you mean to say that we use limits for function that are continuous but here as a,b, c are discrete and cannot take any value because of the equality relation (a+b+c) is not continuous and so we can't say let's look at what happens to k as (a+b+c) approaches 0? Commented Aug 11 at 18:18
• I am saying that the definition of a continuous function requires you to consider a limit everywhere the function is defined, not just at one point. By the way, the sum (not a relation) $a+b+c$ is continuous, but we don’t use that sum when we’re talking about $k$ as a function of three variables. Continuity of a function of multiple variables requires limits of functions of multiple variables, which is a more advanced topic than single-variable limits. Commented Aug 11 at 18:55

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.

• I didn't fully understood your explanation but what I can understand is that are trying to say that limits are defined for functions which are continuous and in my second example a b and c and not continuous i.e they can't have just any value because they have to satisfy the equality relation above? So we can't actually say that as (a+b+c) "approaches ' 0 k approaches 2 because (a+b+c) would not be continuous as a,b and c are not continuous Commented Aug 11 at 12:33
• Not quite. I'm just saying that in your second example it doesn't make much sense to talk about limits because we're dealing with solution sets instead of a single real function. Commented Aug 11 at 21:37
• That's what I meant. Because we are dealing with solution sets, the term (a+b+c) can't have just any value i e we cannot put a ,b and c randomly just like we do in a function because they have to satisfy the relationship as well. So it's not the same as limit of function x/x which can approach 0 and its limit is 1 as it approaches 0. Here (a+b+c) cannot "approach" or get arbitrarily close to 0 it can have only discrete values so it's not a function and so we can't take limits here? Commented Aug 12 at 7:35
• You can put it in several different ways. Once you get to $k(a+b+c)=2(a+b+c)$, you can think of $(a+b+c)$ as a single real variable $x$ and then ask: given a value of $x$, what values of $k$ satisfy the relation? You get a function from reals to sets of reals. In this case $x$ can indeed "approach" 0, as it is just a number, but we don't have a limit because the output of the function is sets. ... Commented Aug 12 at 8:03
• Alternatively, you could ask: given individual values for $a,b,c$, what values of $k$ satisfy the equation? Here you get a function from $\mathbb{R}^3$ to sets of reals. Now $a,b,c$ can each approach a value, or said another way, $(a,b,c)$ can approach a 3D point such as $(1,1,-2)$. But we have the same problem as before to define limits. Finally, you could also just ask: what values of $a,b,c,k$ satisfy the solution? You get a subset of $\mathbb{R}^4$. Here you don't even have a function, like you said, so you can't ask about limits. Commented Aug 12 at 8:08