Mathematical definitions provide logical statements that classify whether or not a property applies or not to a mathematical object. However, in most cases it would quickly become burdensome to verify the logical statement every time, especially if the work is essentially being repeated each time. A theorem provides us with the tool that bypasses the repetitive work and allows us to determine if the property applies or not in a simpler way (called verifying the hypothesis). The proof of the theorem contains the work showing the logical connection between the hypothesis and the conclusion.
The mathematical property of a limit provides a natural illustration of the connection between definitions and theorems. That a function has a limit is the property of interest. The rules for limits consist of a collection of theorems that allow us to construct true limit statements from simpler limits. Once the limit rules are determined, we begin to recognize conditions where limit rules would apply and establish new definitions and theorems (about continuity) that allow us to establish limit statements directly.
The first limit rules give us particular functions for whom we will immediately know the limits. I like to call such rules elementary limits as they provide our building blocks for more complicated algebraic functions. These will be the first collection of limit statements that will be shown to be always true.
Example: Suppose \(f(x) = 3\), \(c=1\) and \(L=3\). We want to see if \[ \lim_{x \to c} f(x) = L \qquad \Leftrightarrow \qquad \lim_{x \to 1} [3] = 3. \]
The function \(f(x)\) is constant, which means that the graph is a horizontal line. For any \(\epsilon > 0\), the inequality \(\big|f(x) - 3\big| < \epsilon\) is equivalent to asking for which values of \(x\) the graph \(y=f(x)\) is between \(y=3-\epsilon\) and \(y=3+\epsilon\). Because \(f(x)=3\) is constant, the solution set is all real numbers. Consequently, every punctured neighborhood \(N(1,\delta)\), with \(\delta > 0\), will be a subset.
The limit statement is true if for every \(\epsilon > 0\) there is at least one \(\delta > 0\) so that \((\epsilon, \delta)\) makes the limit implication true. The values \((\epsilon, \delta)\) so that \(0 < \big|x-1\big| < \delta \) implies \(\big|f(x)-3\big| < \epsilon\) is every value \(\epsilon > 0\) and \(\delta > 0\). So the limit statement is true.
The argument in the example is the same for any value of \(c\) and also for any value \(L\) as long as \(f(x) = L\). This becomes our first limit rule, which we call the limit of a constant.
Theorem (Limit of a Constant Function): For any values \(c\) and \(L\), \[ \lim_{x \to c} [L] = L. \]
Proof:
For the function \(f(x)=L\), the absolute value can be written \(\big|f(x)-L\big| = 0\) so that
\(\big|f(x)-L\big| < \epsilon\) is true for all \(x\). Consequently, for every \(\epsilon > 0\)
and any \(\delta > 0\), the implication
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|f(x)-L\big| < \epsilon \]
is true. Therefore,
\[ \lim_{x \to c} [L] = L. \]
From now on, if we see a constant function and want the corresponding limit, we know the limit of the constant. Instead of going through the \((\epsilon,\delta)\) argument, we just need to refer to our rule.
The second example of an elementary limit rule seems almost too obvious.
Theorem (Limit of the Identity Function): For any values \(c\), \[ \lim_{x \to c} [x] = c. \]
Proof:
For the function \(f(x)=x\) and \(L=c\), the absolute value is simply
\(\big|f(x)-L\big| = |x-c|\) so that
\(\big|f(x)-L\big| < \epsilon\) is the same as saying \(\big|x-c\big| < \epsilon\).
Consequently, for every \(\epsilon > 0\) and \(\delta \le \epsilon\), the implication
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|f(x)-L\big| < \epsilon \]
is true. Therefore,
\[ \lim_{x \to c} [x] = c. \]
The third example of an elementary limit rule is for linear functions.
Example: Suppose \(f(x) = 3x-5\). If we want a limit as \(x \to c=1\), we would expect the limit to be \(L=3(1)-5=-2\). Let us test the limit statement \[ \lim_{x \to c} f(x) = L \qquad \Leftrightarrow \qquad \lim_{x \to 1} [3x-5] = -2. \]
We are interested in the pairs \((\epsilon, \delta)\) so that \[ 0 < \big|x - 1\big| < \delta \quad \Rightarrow \quad \big|(3x-5) - (-2)\big| < \epsilon.\] So we start by finding the solution set to \(\big|3x-5+2\big| = \big|3x-3\big| < \epsilon\). Using the properties of absolute value, we know \[ \big|3x-3\big| = \big|3(x-1)\big| = \big|3\big| \cdot \big|x-1\big| = 3 \big|x-1\big|. \]
Our inequality can then be rewritten in the equivalent forms \[ \big|3x-3\big| < \epsilon \quad \Leftrightarrow \quad 3\big|x-1\big| < \epsilon \quad \Leftrightarrow \quad \big|x-1\big| < {\textstyle \frac{1}{3}} \epsilon. \] The solution set is \(1-\frac{1}{3}\epsilon, 1+\frac{1}{3}\epsilon\). The punctured neighborhood \(N(1,\frac{1}{3}\epsilon)\) is exactly the solution set with the center at \(x=1\) removed.
Consequently, the set of pairs \((\epsilon, \delta)\) for which the limit implication \[ 0 < \big|x-1\big| < \delta \quad \Rightarrow \quad \big|(3x-5) - (-2)\big| < \epsilon\] is true includes \(\delta \le \frac{1}{3} \epsilon\) for all \(\epsilon > 0\). Since each \(\epsilon > 0\) has values \(\delta > 0\) in this relation, the limit statement follows: \[ \lim_{x \to 1} [3x-5] = -2. \]
If we were to consider any other linear function, the argument would be essentially identical but with different numbers. Consequently, if we generalize each number using a variable, we obtain a corresponding limit statement for any linear function.
Theorem (Limit of a Linear Function): For any values \(m\), \(b\) and \(c\), \[ \lim_{x \to c} [mx+b] = mc+b. \]
Proof:
We are needing to prove the limit statement for \(f(x) = mx+b\) and \(L=mc+b\).
When \(m=0\), \(f(x) = b\) is a constant function, which was an earlier rule.
So for the rest of the proof, we assume \(m \ne 0\).
The inequality \(\big|f(x)-L\big| < \epsilon\) can be rewritten
\[
\begin{align*}
\big|f(x)-L\big| < \epsilon \quad & \Leftrightarrow \quad \big|(mx+b) - (mc+b)\big| < \epsilon \\
\quad & \Leftrightarrow \quad \big|m(x-c)\big| < \epsilon \\
\quad & \Leftrightarrow \quad \big|m\big| \cdot \big|x-c\big| < \epsilon \\
\quad & \Leftrightarrow \quad \big|x-c\big| < {\textstyle \frac{1}{\big|m\big|}} \epsilon
\end{align*}
\]
So for every \(\epsilon > 0\), the values \(\delta \le \frac{1}{\big|m\big|} \epsilon\) will make the implication \[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|f(x) - L\big| < \epsilon\] true. Since for each \(\epsilon > 0\), there is a value \(\delta > 0\) making the implication true, the limit statement is true, \[ \lim_{x \to c} [mx+b] = mc+b. \]
Now that we know the limits of any constant or linear function, we can start to build more complicated functions by using arithmetic. The simplest operations involve a single function, say \(f(x)\), using a constant sum, \(f(x) + k\), a constant multiple, \(k \cdot f(x)\), or a reciprocal, \(\div f(x) = 1/f(x)\). (multiplicative inverse). (The additive inverse \(-f(x)\) is just a constant multiple.)
Each theorem has a corresponding proof. The proof for the reciprocal is significantly more challenging than the other two theorems' proofs.
Theorem (Constant Sum Rule): If \(\displaystyle \lim_{x \to c} f(x) = L\) and \(k\) is any number, then \[ \lim_{x \to c} [f(x) + k] = L + k. \]
Proof:
Because \(\lim_{x \to c} f(x) = L\), we know that for any value \(\epsilon > 0\),
there will be values \(\delta > 0\) so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|f(x) - L\big| < \epsilon. \]
We wish to show that the limit statement
\[ \lim_{x \to c} [f(x) + k] = L+k, \]
so we need to show that for any value \(\epsilon > 0\), there is a value \(\delta > 0\)
so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|(f(x)+k) - (L+k)\big| < \epsilon. \]
Because \(\big|(f(x)+k) - (L+k)\big| = \big|f(x) - L\big|\), the same \((\epsilon, \delta)\)
pairs that satisfy the limit implication for \(f(x)\), \(c\) and limit \(L\)
will satisfy the limit implication for \(f(x)+k\), \(c\) and limit \(L+k\).
Therefore,
\[ \lim_{x \to c} [f(x) + k] = L + k. \]
Theorem (Constant Multiple Rule): If \(\displaystyle \lim_{x \to c} f(x) = L\) and \(k\) is any number, then \[ \lim_{x \to c} [k \cdot f(x)] = k \cdot L. \]
Proof:
If \(k=0\), then \(k \cdot f(x) = 0\) is a constant functions, which has an elementary
limit, \(\lim_{x \to c}[0] = 0\). Since \(0 \cdot L = 0\), this case is true.
For the remainder of the proof, we assume \(k \ne 0\). Because \(\lim_{x \to c} f(x) = L\), we know that for any value \(\epsilon > 0\), there will be values \(\delta > 0\) so that \[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|f(x) - L\big| < \epsilon. \] We wish to show that the limit statement \[ \lim_{x \to c} [k \cdot f(x)] = k \cdot L, \] so we need to show that for any value \(\epsilon > 0\), there is a value \(\delta > 0\) so that \[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|k \cdot f(x) - k \cdot L\big| < \epsilon. \] Because \(\big|k \cdot f(x) - k \cdot L\big| = \big|k\big|\big|f(x) - L\big|\), the following inequalities are equivalent, \[ \big|k \cdot f(x) - k \cdot L\big| < \epsilon \quad \Leftrightarrow \quad \big|f(x) - L\big| < \frac{1}{\big|k\big|} \epsilon.\] Since \(\frac{1}{\big|k\big|} \epsilon > 0\) and \(\displaystyle \lim_{x \to c} f(x) = L\), there is a number \(\delta > 0\) so that \[ \begin{align*} 0 < \big|x-c\big| < \delta \quad &\Rightarrow \quad \big|f(x) - L\big| < \frac{1}{\big|k\big|}\epsilon \\ & \Rightarrow \quad \big|k \cdot f(x) - k \cdot L\big| < \epsilon. \end{align*} \] Therefore, \[ \lim_{x \to c} [k \cdot f(x)] = k \cdot L. \]
Theorem (Inverse Rule or Reciprocal Rule): If \(\displaystyle \lim_{x \to c} f(x) = L\) and \(L \ne 0\), then \[ \lim_{x \to c} [\div f(x)] = \div L \quad \hbox{also written} \quad \lim_{x \to c} \frac{1}{f(x)} = \frac{1}{L}. \]
Proof:
The hypothesis is that \(\displaystyle \lim_{x \to c} f(x) = L\).
So for any value \(\epsilon > 0\), there will be values \(\delta > 0\) so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|f(x) - L\big| < \epsilon. \]
Because this theorem is about the reciprocal, we need \(f(x)\) to stay away from 0. Using a value \(\epsilon = \frac{1}{2}\big|L\big| > 0\), we can identify a value \(\delta_1 > 0\) so that \(0 < \big|x-c\big| < \delta_1\) implies \(\big|f(x) - L\big| < \frac{1}{2}\big|L\big|\), which in turn implies \(\big|f(x)\big| > \frac{1}{2}\big|L\big|\) and \(\displaystyle \frac{1}{\big|f(x)\big|} < \frac{2}{\big|L\big|}\).
We now consider the inequality relevant to the limit of interest. Given any number \(\epsilon > 0\), we consider the inequality \[ \left| \frac{1}{f(x)} - \frac{1}{L} \right| < \epsilon. \] Finding a common denominator, this is equivalent to \[ \left| \frac{L - f(x)}{L \cdot f(x)} \right| < \epsilon \quad \Leftrightarrow \quad \frac{1}{\big|L\big|\big|f(x)\big|} \cdot \big|f(x)-L\big| < \epsilon \]
So long as \(0 < \big|x-c\big| < \delta_1\), we know \(\displaystyle \frac{1}{\big|L\big|\big|f(x)\big|} < \frac{2}{L^2}\). Since \(\frac{L^2}{2} \epsilon > 0\), we can find another value \(\delta_2 > 0\) so that \[ 0 < \big|x-c\big| < \delta_2 \quad \Rightarrow \quad \big|f(x) - L\big| < \frac{L^2}{2} \epsilon. \] Using \(\delta = \min(\delta_1, \delta_2)\), if \(0 < \big|x-c\big| < \delta\) then both hypotheses are satisfied so that \(\displaystyle \frac{1}{\big|L\big|\big|f(x)\big|} < \frac{2}{L^2}\) and \(\big|f(x)-L\big| < \frac{L^2}{2} \epsilon\). Properties of inequalities then lead to \[ \frac{1}{\big|L\big|\big|f(x)\big|} \cdot \big|f(x)-L\big| < \frac{2}{L^2} \cdot \frac{L^2}{2} \epsilon = \epsilon. \]
Since every \(\epsilon > 0\) has a corresponding \(\delta > 0\) so that \[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \left| \frac{1}{f(x)} - \frac{1}{L} \right| < \epsilon, \] we have proved \[ \lim_{x \to c} \frac{1}{f(x)} = \frac{1}{L}. \]
Each of these rules include a hypothesis that we already know the limit statement \[ \lim_{x \to c} f(x) = L \] is true. The limit rules allow us to conclude new limit statements are true.
Example: Suppose that \(f(x)\) is any function such that \(\displaystyle \lim_{x \to 2} f(x) = 5\). This example will illustrate each single-function rule.
Using the value \(k=3\), the constant sum rule for limits guarantees that \[ \lim_{x \to 2} [f(x) + 3] = 5+3 = 8. \] Using the same value \(k=3\), the constant multiple rule for limits guarantees \[ \lim_{x \to 2} [3 f(x) ] = 3 \cdot 5 = 15. \] Finally, the reciprocal rule for limits guarantees \[ \lim_{x \to 2} \frac{1}{f(x)} = \frac{1}{5}. \] The limit rules, being theorems, guarantee these limit statements are true to the same amount of logical certainty as if we had applied the definition itself.
The next group of limit rules involve the arithmetic of two functions. In each theorem, the hypothesis states that the limit of two functions are known for the same \(x\)-value \(c\). The conclusion of each theorem provides a true limit statement involving an arithmetic combination of the two functions. The proof for the theorem for the limit of a product is the most challenging. The proofs for a difference and a quotient are the easiest, because they illustrate how to join different theorems to reach a conclusion.
Theorem (Sum Rule): If \(\displaystyle \lim_{x \to c} f(x) = L\) and \(\displaystyle \lim_{x \to c} g(x) = M\), then \[ \lim_{x \to c} [f(x) + g(x)] = L + M. \]
Proof:
Because \(\lim_{x \to c} f(x) = L\), we know that for any value \(\epsilon > 0\),
there will be values \(\delta > 0\) so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|f(x) - L\big| < \epsilon. \]
Similarly, because \(\lim_{x \to c} g(x) = M\), we know that for any value \(\epsilon > 0\),
there will be values \(\delta > 0\) so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|g(x) - M\big| < \epsilon. \]
We wish to show that the limit statement
\[ \lim_{x \to c} [f(x) + g(x)] = L + M, \]
so we need to show that for any value \(\epsilon > 0\), there is a value \(\delta > 0\)
so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|(f(x)+g(x) - (L+M)\big| < \epsilon. \]
Rewriting the absolute value and then using the triangle inequality, we find \[ \big|(f(x)+g(x)) - (L+M)\big| = \big|(f(x)-L) + (g(x)-M)\big| \le \big|f(x)-L\big| + \big|g(x)-M\big|. \] The triangle inequality tells us that we need to make both \(\big|f(x)-L\big|\) and \(\big|g(x)-M\big|\) small enough so that the sum is less than \(\epsilon\). If we can guarantee that each of these are less than \(\frac{1}{2}\epsilon\) individually, then the sum will be less than \(\epsilon\).
Since \(\frac{1}{2}\epsilon > 0\), the limit for \(f(x)\) guarantees that we can find \(\delta_1 > 0\) so that \[ 0 < \big|x-c\big| < \delta_1 \quad \Rightarrow \quad \big|f(x) - L\big| < \frac{1}{2} \epsilon. \] Similarly, the limit for \(g(x)\) guarantees that we can find \(\delta_2 > 0\) so that \[ 0 < \big|x-c\big| < \delta_2 \quad \Rightarrow \quad \big|g(x) - M\big| < \frac{1}{2} \epsilon. \] We let \(\delta = \min(\delta_1, \delta_2\). If \(0 < \big|x-c\big| < \delta\), then \(\big|f(x)-L\big| < \frac{1}{2}\epsilon\) and \(\big|g(x)-M\big| < \frac{1}{2}\epsilon\). Consequently, \[ \big|(f(x)+g(x)) - (L+M)\big| \le \big|f(x)-L\big| + \big|g(x)-M\big| < \epsilon. \] Therefore, \[ \lim_{x \to c} [f(x)+g(x)] = L+M. \]
Theorem (Difference Rule): If \(\displaystyle \lim_{x \to c} f(x) = L\) and \(\displaystyle \lim_{x \to c} g(x) = M\), then \[ \lim_{x \to c} [f(x) - g(x)] = L - M. \]
Proof:
The difference is really just the sum using an additive inverse.
By hypothesis, we know \(\displaystyle \lim_{x \to c} f(x) = L\).
Because we also know \(\displaystyle \lim_{x \to c} g(x) = M\), using the limit of a
constant multiple (with \(k=-1\)), we know \(\displaystyle \lim_{x \to c} -g(x) = -M\).
Then using the limit of a sum, we know
\[ \lim_{x \to c} [f(x) - g(x)] = \lim_{x \to c} [f(x) + -g(x)] = L + -M = L-M. \]
Theorem (Product Rule): If \(\displaystyle \lim_{x \to c} f(x) = L\) and \(\displaystyle \lim_{x \to c} g(x) = M\), then \[ \lim_{x \to c} [f(x) + g(x)] = L + M. \]
Proof:
Because \(\displaystyle \lim_{x \to c} f(x) = L\), we know that for any value \(\epsilon > 0\),
there will be values \(\delta > 0\) so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|f(x) - L\big| < \epsilon. \]
Similarly, because \(\displaystyle \lim_{x \to c} g(x) = M\), we know that for any value \(\epsilon > 0\),
there will be values \(\delta > 0\) so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|g(x) - M\big| < \epsilon. \]
We wish to show that the limit statement
\[ \lim_{x \to c} [f(x) \cdot g(x)] = L \cdot M, \]
so we need to show that for any value \(\epsilon > 0\), there is a value \(\delta > 0\)
so that
\[ 0 < \big|x-c\big| < \delta \quad \Rightarrow \quad \big|(f(x) \cdot g(x) - (L \cdot M)\big| < \epsilon. \]
This theorem uses the common algebraic strategy of adding inverses in order to regroup the terms in the absolute value, this time adding \(-f(x) \cdot M\) and \(f(x) \cdot M\). This allows us to rewrite the absolute value in the implication as \[ \begin{align*} \big|f(x) \cdot g(x) - L \cdot M\big| &= \big|f(x) \cdot g(x) - f(x) \cdot M + f(x) \cdot M - L \cdot M\big| \\ &= \big|f(x)(g(x)-M) + M(f(x)-L)\big|. \end{align*} \] By the triangle inequality, we know that \[ \big|f(x) \cdot g(x) - L \cdot M\big| \le \big|f(x)\big| \cdot \big|g(x)-M\big| + \big|M\big| \cdot \big|f(x)-L\big|. \]
Using the limit for \(f(x)\), we can choose \(\delta_1\) so that \(\big|f(x) - L\big| < 1\) whenever \(0 < \big|x-c\big| < \delta_1\). This will then guarantee that \(\big|f(x)\big| < \big|L\big|+1\). We can also choose \(\delta_2\) so that \[\big|f(x) - L\big| < \frac{1}{\big|M\big|+1}\frac{\epsilon}{2} \] whenever \(0 < \big|x-c\big| < \delta_2\). Finally, we can choose \(\delta_3\) so that \[ \big|g(x)-M\big| < \frac{1}{\big|L\big|+1} \frac{\epsilon}{2} \] whenever \(0 < \big|x-c\big| < \delta_3\). We let \(\delta = \min(\delta_1, \delta_2, \delta_3\). If \(0 < \big|x-c\big| < \delta\), all three hypotheses are satisfied and we know \[ \begin{align*} \big|f(x)\big| &< \big|L\big| + 1 \\ \big|g(x) - M\big| &< \frac{1}{\big|L\big|+1} \frac{\epsilon}{2} \\ \big|f(x) - L\big| &< \frac{1}{\big|M\big|+1} \frac{\epsilon}{2}. \end{align*} \] Consequently, \[ \begin{align*} \big|(f(x)\cdot g(x)) - (L \cdot M)\big| &\le \big|f(x)\big|\big|g(x)-M\big| + \big|M\big|\big|f(x)-L\big| \\ &< (\big|L\big|+1) \cdot \frac{1}{\big|L\big|+1}\frac{\epsilon}{2} + \big|M\big| \cdot \frac{1}{\big|M\big|+1}\frac{\epsilon}{2} \\ &< \epsilon. \end{align*} \] That is, we have shown that for every \(\epsilon > 0\), there are values \(\delta > 0\) so that \(0 < \big|x-c\big| < \delta\) implies \(\big|f(x)\cdot g(x) - L \cdot M \big| < \epsilon\). Therefore, \[ \lim_{x \to c} [f(x) \cdot g(x)] = L \cdot M. \]
Theorem (Quotient Rule): If \(\displaystyle \lim_{x \to c} f(x) = L\) and \(\displaystyle \lim_{x \to c} g(x) = M\) and \(M \ne 0\), then \[ \lim_{x \to c} [f(x) \div g(x)] = L \div M. \]
Proof:
The quotient is really just the product using a multiplicative inverse (reciprocal).
By hypothesis, we know \(\displaystyle \lim_{x \to c} f(x) = L\).
We also know \(\displaystyle \lim_{x \to c} g(x) = M\) and \(M \ne 0\),
so we can use the limit of a reciprocal to know
\[ \lim_{x \to c} \div g(x) = \lim_{x \to c} \frac{1}{g(x)} = \frac{1}{M} = \div M. \]
Then using the limit of a product, we know
\[ \lim_{x \to c} [f(x) \div g(x)] = \lim_{x \to c} [f(x) \cdot \frac{1}{g(x)}]
= L \cdot \frac{1}{M} = L \div M. \]
Relatively complex algebraic functions can be constructed using only basic arithmetic operations on the basic building blocks of constants and linear functions. Every positive integer power of \(x\) can be constructed by repeatedly multiplying by the identity function \(i(x) = x\). Adding constant multiples of powers gives us any polynomial. The quotient of polynomials gives us rational functions.
By recognizing how a function is constructed, and then using the corresponding rules for limits, we can determine the limit of any function that can be formed from elementary building blocks and the basic arithmetic operations.
Example: Determine the appropriate limit statement for \(\displaystyle \lim_{x \to 3} \frac{x^2-2x}{4(x^2+5)}\) using limit rules. Organize the statements in the form of a proof.
So that the statements are organized in the form of a proof, we need to be sure that each time we apply a rule, we have previously showed the limits of the component parts of the new formula. The first few statements should be elementary limits, and the remainder will be combinations.
Each statement in this sequence for the previous example was justified using the limit theorems and earlier steps in the sequence. The last line in the sequence gives the limit of interest. Using limit rules in this way is actually just as much a proof of the limit as if we were to figure out a careful argument using \((\epsilon, \delta)\) and the limit definition.
Hopefully you agree that using the limit rules is much easier, as they allow us to focus on higher-level structure. Good definitions and theorems in mathematics enable and empower our understanding in a similar way.
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