##### Theorem5.7.1Limit of a Constant (LE:Const)

Hypothesis: \(k\) is a real number.

Conclusion: \(\displaystyle \lim_{x \to a} k = k\).

So far, we have been taking advantage of the fact that simple algebraic formulas are continuous so that when taking limits all we do is use the value of the formula at the point in question. However, this is not always the case.

In principle, to find a limit \(\displaystyle \lim_{x \to a} f(x)\), you would need to create a table and evaluate at a sequence of values that is converging to \(a\) and see what the sequence of values \(f(x)\) is converging toward. (And that value should not depend on which sequence you tried.) What is it that allows us to just use a formula instead of using tables every time we need to find a limit? The answer is limit rules.

Mathematics uses deductive reasoning (built on axioms and logical implications) to establish rules for what must be true based on our fundamental assumptions of how things work. These proven consequences are called theorems. Limit rules are a powerful collection of theorems that allow us to find limits in ways that do not require tables, but allow us to build up a collection of known limits that we can use in more complicated problems.

Every theorem has two parts—a hypothesis and a conclusion. A hypothesis is a collection of conditions that must be checked to see if the theorem's conclusion can be applied. A conclusion is a statement that is guaranteed to be true whenever the hypothesis has been satisfied. Whenever, we wish to apply a theorem, it is our duty to first check that all of the conditions in the hypothesis have been satisfied. Once this is complete, then you interpret the conclusion for the context of your problem while citing the theorem as your justification.

The first collection of limit rules are some basic limits. We can think of them as our building blocks for more complicated limits. In reality, each of these limit rules are really statements about continuity of these very elementary functions.

Hypothesis: \(k\) is a real number.

Conclusion: \(\displaystyle \lim_{x \to a} k = k\).

Hypothesis: none

Conclusion: \(\displaystyle \lim_{x \to a} x = a\).

Note: The name of the previous theorem is based on the function \(f(x)=x\) being called the identity function because the output is exactly the input.

Hypothesis: \(m\) and \(b\) are real numbers.

Conclusion: \(\displaystyle \lim_{x \to a} [mx+b] = ma+b\).

The second collection of limit rules tell us how we can take limits that we already know (starting with building blocks) and use them to compute more complicated limits. The first three rules take a single limit that is known to be valid and use arithmetic with a constant to find a new limit.

Hypothesis: \(\displaystyle \lim_{x \to a} f(x)=L\) and \(k\) is a real number.

Conclusion: \(\displaystyle \lim_{x \to a} [f(x)+k] = L+k\).

Hypothesis: \(\displaystyle \lim_{x \to a} f(x)=L\) and \(k\) is a real number.

Conclusion: \(\displaystyle \lim_{x \to a} [k \cdot f(x)] = k \cdot L\).

Hypothesis: \(\displaystyle \lim_{x \to a} f(x)=L\) and \(L \ne 0\).

Conclusion: \(\displaystyle \lim_{x \to a} \frac{1}{f(x)} = \frac{1}{L}\).

The next limit rules of combination allow us to take two limits that we know and combine them with arithmetic. In each of the cases, note that both limits in the hypothesis have \(x \to a\) (i.e., \(x\) approaches the same value in both limits).

Hypothesis: \(\displaystyle \lim_{x \to a} f(x)=L\) and \(\displaystyle \lim_{x \to a} g(x)=M\).

Conclusion: \(\displaystyle \lim_{x \to a} [f(x)+g(x)] = L+M\).

Hypothesis: \(\displaystyle \lim_{x \to a} f(x)=L\) and \(\displaystyle \lim_{x \to a} g(x)=M\).

Conclusion: \(\displaystyle \lim_{x \to a} [f(x)-g(x)] = L-M\).

Hypothesis: \(\displaystyle \lim_{x \to a} f(x)=L\) and \(\displaystyle \lim_{x \to a} g(x)=M\).

Conclusion: \(\displaystyle \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M\).

Hypothesis: \(\displaystyle \lim_{x \to a} f(x)=L\) and \(\displaystyle \lim_{x \to a} g(x)=M\) and \(M \ne 0\).

Conclusion: \(\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}\).

In addition to the arithmetic of functions, composition of functions plays an important role in algebra and calculus. So we need a limit rule associated with composition. Recall that for composition, the output of one function becomes the input to another function.

Hypothesis: \(\displaystyle \lim_{x \to a} f(x)=L\) and \(g\) is continuous at \(L\), or in other words, \(\displaystyle \lim_{u \to L} g(u) = g(L)\).

Conclusion: \(\displaystyle \lim_{x \to a} g \circ f(x) = \lim_{x \to a} g(f(x)) = g(L)\).

If \(f\) is continuous at \(a\) so that \(\displaystyle \lim_{x \to a} f(x) = f(a)\) then we have \(\displaystyle \lim_{x \to a} g \circ f(x) = g \circ f(a)\). In other words, the composition of continuous functions is a continuous function.

Each of these limit rules is proved using a more precise definition of what a limit means. For now, we will consider these rules as a starting point for evaluating limits.

There are two ways in which limit rules are applied. One way is to provide formal justification of limit calculations, or in other words, to write a proof of limit statements. The other way limit rules are used is to break a computation down into recognizable and manageable parts. This section focuses on the process of formal justification.

A mathematical proof is essentially a sequence of statements, each of which is demonstrably true based only on previously stated knowledge and logical arguments. This means that when writing a proof or any careful justification, we must be careful that when we write something down we have previously established all of the necessary conditions at a previous step. In order to avoid circular reasoning, we should avoid referring to something as true before we actually show it is true.

For justification of limit statements, this means that we start from the small building blocks that create our formula and put them together one step at a time until we can justify the limit statement we are trying to prove.

Compute and justify \(\displaystyle \lim_{x \to 2} 3x^2(2x-5)\).

Compute and justify \(\displaystyle \lim_{x \to 3} x^3+4x^2-3x+1\).