Properties of Limits
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.3: “The Limit Laws” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws |
| Supplementary | OpenStax Calculus Volume 1, Section 2.2: “The Limit of a Function” |
| Supplementary link | https://openstax.org/books/calculus-volume-1/pages/2-2-the-limit-of-a-function |
| Textbook used in class | Stewart, Calculus, Section 1.6: “Calculating Limits Using the Limit Laws” (Examples 2, 3, 5, 6, 11) |
Both OpenStax sources are free and openly licensed.
Overview
The limit laws are the rules that let you compute a limit without guessing from a table or a graph. They say one thing: limits respect arithmetic.
If you know what $f(x)$ approaches and what $g(x)$ approaches, then $f(x) + g(x)$ approaches the sum, $f(x)g(x)$ approaches the product, and so on. The limit of a combination is the combination of the limits. That is what the laws state, and it is exactly what you would hope is true.
This leads to a shortcut that handles most limits in one step. For a polynomial or a rational function, the limit as $x \to a$ is just the value $f(a)$, found by substituting. That is called the Direct Substitution Property. A polynomial is built out of constants and copies of $x$ with arithmetic, so applying the laws to each operation lands you on $f(a)$.
The interesting work begins where substitution fails. When plugging in gives the meaningless form $\frac{0}{0}$, the laws do not apply directly and you need a repair: rewrite the function into one that agrees with the original everywhere except the trouble point, then substitute into the rewrite. Three repairs cover nearly everything: factor and cancel, rationalize with a conjugate, or trap the function between two others with the Squeeze Theorem. Knowing which repair to reach for is the real skill of this section.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
If one-sided limits are unfamiliar, review them first. Several limits in this section exist only because their one-sided limits agree, or fail to exist because they disagree.
Quick Reference
The Limit Laws. Suppose $c$ is a constant and $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist. Then:
| Law | Statement |
|---|---|
| Sum | $\lim_{x\to a}[f(x) + g(x)] = \lim_{x\to a} f(x) + \lim_{x\to a} g(x)$ |
| Difference | $\lim_{x\to a}[f(x) - g(x)] = \lim_{x\to a} f(x) - \lim_{x\to a} g(x)$ |
| Constant multiple | $\lim_{x\to a}[c\,f(x)] = c\lim_{x\to a} f(x)$ |
| Product | $\lim_{x\to a}[f(x)g(x)] = \lim_{x\to a} f(x)\cdot\lim_{x\to a} g(x)$ |
| Quotient | $\lim_{x\to a}\dfrac{f(x)}{g(x)} = \dfrac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)}$, provided $\lim_{x\to a} g(x) \neq 0$ |
| Power | $\lim_{x\to a}[f(x)]^n = \big[\lim_{x\to a} f(x)\big]^n$ |
| Root | $\lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to a} f(x)}$ (if $n$ is even, need the inside limit $> 0$) |
Two special limits. $\lim_{x\to a} c = c$ and $\lim_{x\to a} x = a$.
Direct Substitution Property. If $f$ is a polynomial or a rational function and $a$ is in the domain of $f$, then $\lim_{x\to a} f(x) = f(a)$.
When substitution gives $\frac{0}{0}$, use one of three repairs:
- Factor and cancel the common factor, then substitute.
- Rationalize with a conjugate (for limits involving roots), then substitute.
- Squeeze the function between two others with the same limit.
Key Concepts
1. The Laws and the Direct Substitution Shortcut
The laws say limits pass through arithmetic. Combined with the two special limits $\lim_{x\to a} c = c$ and $\lim_{x\to a} x = a$, they let you evaluate any polynomial or rational limit by substitution.
Example 1. Evaluate the limits, justifying each step. (This is Stewart 1.6, Example 2.)
(a) $\;\displaystyle\lim_{x\to 5}(2x^2 - 3x + 4)$
By the sum and difference laws, then the constant-multiple and power laws: \[ \lim_{x\to 5}(2x^2 - 3x + 4) = 2\lim_{x\to 5} x^2 - 3\lim_{x\to 5} x + \lim_{x\to 5} 4 = 2(5^2) - 3(5) + 4 = 39. \]
(b) $\;\displaystyle\lim_{x\to -2}\frac{x^3 + 2x^2 - 1}{5 - 3x}$
The denominator limit is $5 - 3(-2) = 11 \neq 0$, so the quotient law applies: \[ \lim_{x\to -2}\frac{x^3 + 2x^2 - 1}{5 - 3x} = \frac{(-2)^3 + 2(-2)^2 - 1}{5 - 3(-2)} = \frac{-8 + 8 - 1}{11} = -\frac{1}{11}. \]
Boxed answers: $39$ and $-\dfrac{1}{11}$.
Recap. In both parts the final answer is exactly what substitution gives, because the function is a polynomial (a) or a rational function whose denominator is nonzero at the point (b). This is the Direct Substitution Property: for such functions, $\lim_{x\to a} f(x) = f(a)$. The laws are what make the shortcut legal.
2. When Substitution Fails: the $\frac{0}{0}$ Form
Direct substitution stops working when the denominator approaches zero. If the numerator also approaches zero, the result is the indeterminate form $\frac{0}{0}$, which carries no information by itself; the limit might be any number, or might not exist. The repair is to rewrite the function as a simpler one that agrees with it away from the trouble point.
The justification is a small but important fact: if $f(x) = g(x)$ for all $x \neq a$, then $\lim_{x\to a} f(x) = \lim_{x\to a} g(x)$. A limit as $x \to a$ never looks at the value exactly at $a$, only nearby. So replacing $f$ with a function that agrees everywhere except at $a$ leaves the limit unchanged.
Important: $\frac{0}{0}$ is a signal to do algebra. It carries no value of its own. Writing “$\frac00 = 1$” or “$\frac00 = 0$” is wrong. The form $\frac00$ means the quotient law does not apply and you must rewrite the expression first. Cancel a common factor, rationalize, or squeeze, and then evaluate.
“the limit is what happens as you get close but never touch $a$, so you can never know the exact answer.”
This is the limit-as-unreachable-barrier error. The limit $\lim_{x \to a} f(x) = L$ is a precise, exact statement. It does not mean the function is “almost $L$” or that $x$ is somehow stuck before $a$. The statement says: given any desired level of closeness to $L$, there is a corresponding closeness to $a$ that guarantees it. The output gets as close to $L$ as you demand, and the limit IS the value $L$, not an approximation to it. When the limit laws are used to compute $\lim_{x \to 5}(2x^2 - 3x + 4) = 39$, that 39 is exact.
A specific example that breaks the barrier image: the limit $\lim_{x \to 1} (x + 1) = 2$ holds because $x + 1$ approaches 2 as $x$ approaches 1. There is no wall at $x = 1$; the function simply evaluates to 2 there as well, because it is continuous. The limit is exactly 2, not “just under 2” or “approaching 2 but never arriving.”
limits are only about moving objects or physical motion.
This is the limit-as-only-dynamic-motion error. The notation “$x \to a$” looks like motion, and velocity problems provide a natural first context. A limit describes the behavior of a function near a point. Nothing physical has to be moving. The limit $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$ concerns the algebraic function $(x^2 - 4)/(x - 2)$ and what value it approaches near $x = 2$. No object is moving. The dynamic language (“as $x$ moves toward $a$”) is a reading aid, not a physical claim. Limits apply equally to functions defined by tables, graphs, or formulas with no physical interpretation.
3. Repair 1: Factor and Cancel
When the numerator and denominator share a factor that causes the $\frac{0}{0}$, cancel it.
Example 2. Find $\displaystyle\lim_{x\to 1}\frac{x^2 - 1}{x - 1}$. (This is Stewart 1.6, Example 3.)
Goal. Substitution gives $\frac{0}{0}$. Factor the numerator, cancel, then substitute.
Factor the difference of squares: \[ \frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1}. \] For $x \neq 1$ the common factor $x - 1$ is nonzero, so it cancels: \[ \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad (x \neq 1). \] The simplified function $x + 1$ agrees with the original everywhere except $x = 1$, so they have the same limit: \[ \lim_{x\to 1}\frac{x^2 - 1}{x - 1} = \lim_{x\to 1}(x + 1) = 2. \]
Boxed answer: $2$.
Recap. The original function is undefined at $x = 1$ (a hole in the graph), but the limit still exists, because the curve heads toward $2$ from both sides. Canceling is legal precisely because the limit ignores the single point $x = 1$.
4. Repair 1 Applied to a Difference Quotient
Difference quotients always produce $\frac{0}{0}$ as $h \to 0$, and factoring out the $h$ is the repair. This is the computation behind every derivative.
Example 3. Evaluate $\displaystyle\lim_{h\to 0}\frac{(3+h)^2 - 9}{h}$. (This is Stewart 1.6, Example 5.)
Goal. Expand the numerator, cancel the $h$, then substitute.
Expand and simplify: \[ \frac{(3+h)^2 - 9}{h} = \frac{9 + 6h + h^2 - 9}{h} = \frac{6h + h^2}{h} = \frac{h(6 + h)}{h} = 6 + h \quad (h \neq 0). \] Now substitute: \[ \lim_{h\to 0}\frac{(3+h)^2 - 9}{h} = \lim_{h\to 0}(6 + h) = 6. \]
Boxed answer: $6$.
Recap. This is the derivative of $x^2$ at $x = 3$ in disguise: it equals $2(3) = 6$. Every derivative you compute from the definition is a $\frac00$ limit repaired by canceling $h$.
5. Repair 2: Rationalize with a Conjugate
When a square root produces the $\frac00$, multiplying by the conjugate clears the root and exposes a cancelable factor.
Example 4. Find $\displaystyle\lim_{t\to 0}\frac{\sqrt{t^2 + 9} - 3}{t^2}$. (This is Stewart 1.6, Example 6.)
Goal. Substitution gives $\frac00$. Multiply numerator and denominator by the conjugate $\sqrt{t^2+9} + 3$.
\[ \frac{\sqrt{t^2 + 9} - 3}{t^2}\cdot\frac{\sqrt{t^2 + 9} + 3}{\sqrt{t^2 + 9} + 3} = \frac{(t^2 + 9) - 9}{t^2\left(\sqrt{t^2 + 9} + 3\right)} = \frac{t^2}{t^2\left(\sqrt{t^2 + 9} + 3\right)}. \] Cancel the $t^2$: \[ = \frac{1}{\sqrt{t^2 + 9} + 3}. \] Now substitute $t = 0$: \[ \lim_{t\to 0}\frac{\sqrt{t^2 + 9} - 3}{t^2} = \frac{1}{\sqrt{0 + 9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}. \]
Boxed answer: $\dfrac{1}{6}$.
Recap. The conjugate turns a difference of a root and a number into a difference of squares on top, which cancels the troublesome factor. Reach for it whenever a root sits next to the $\frac00$.
6. One-Sided Limits and the Existence Theorem
Some limits must be checked from each side. A two-sided limit exists if and only if both one-sided limits exist and are equal.
Example 5. Determine whether $\displaystyle\lim_{x\to 0}\frac{|x|}{x}$ exists. (This is Stewart 1.6, Example 8.)
For $x > 0$, $|x| = x$, so $\dfrac{|x|}{x} = 1$ and $\displaystyle\lim_{x\to 0^+}\frac{|x|}{x} = 1$.
For $x < 0$, $|x| = -x$, so $\dfrac{|x|}{x} = -1$ and $\displaystyle\lim_{x\to 0^-}\frac{|x|}{x} = -1$.
The one-sided limits disagree ($1 \neq -1$), so the two-sided limit does not exist.
Boxed answer: $\displaystyle\lim_{x\to 0}\frac{|x|}{x}$ does not exist.
Recap. When a function is defined by different rules on each side of the target (absolute value, piecewise functions, the greatest-integer function), check both one-sided limits. Equal means the limit exists; unequal means it does not.
7. Repair 3: The Squeeze Theorem
Some limits resist algebra because a factor oscillates and has no limit of its own. The Squeeze Theorem handles these by trapping the function between two others that share a limit.
The Squeeze Theorem. If $f(x) \leq g(x) \leq h(x)$ for $x$ near $a$ (except possibly at $a$), and $\displaystyle\lim_{x\to a} f(x) = \lim_{x\to a} h(x) = L$, then $\displaystyle\lim_{x\to a} g(x) = L$. (Plain gloss: a function pinned between two others that meet at $L$ is forced to $L$ as well.)
Example 6. Show that $\displaystyle\lim_{x\to 0} x^2 \sin\frac{1}{x} = 0$. (This is Stewart 1.6, Example 11.)
Goal. The product law fails because $\sin(1/x)$ has no limit at $0$. Bound the function instead.
Since $-1 \leq \sin\frac{1}{x} \leq 1$ for every $x \neq 0$, and $x^2 \geq 0$, multiply through by $x^2$: \[ -x^2 \leq x^2\sin\frac{1}{x} \leq x^2. \] Both outer bounds go to zero: $\displaystyle\lim_{x\to 0} x^2 = 0$ and $\displaystyle\lim_{x\to 0}(-x^2) = 0$. By the Squeeze Theorem, \[ \lim_{x\to 0} x^2\sin\frac{1}{x} = 0. \]
Boxed answer: $0$.
Recap. The Squeeze Theorem is the tool of last resort when a bounded but wildly oscillating factor blocks the product law. Find a smaller function and a bigger function with the same limit, and the one in the middle is trapped.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using the quotient law when the denominator limit is $0$ | $\lim_{x\to 1}\frac{x^2-1}{x-1} = \frac{0}{0}$ | The law requires a nonzero denominator limit; factor and cancel first |
| Treating $\frac00$ as a number | “$\frac00 = 1$” | It is indeterminate; rewrite the expression, then evaluate |
| Forgetting to check both sides | reporting $\lim_{x\to 0}\frac{|x|}{x} = 1$ | The left limit is $-1$; the two-sided limit does not exist |
| Applying the product law to an oscillating factor | $\lim_{x\to 0} x^2\sin\frac1x = 0 \cdot (\text{DNE})$ | $\sin(1/x)$ has no limit; use the Squeeze Theorem |
| Forgetting the even-root condition | $\lim \sqrt{f(x)}$ with the inside limit negative | The root law needs the inside limit positive for even roots |
| Canceling before checking $x \neq a$ | silently dropping a factor that is zero at $a$ | Cancellation is valid only for $x \neq a$; that is exactly why the limit ignores $a$ |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Evaluate $\displaystyle\lim_{x\to 3}(4x^2 - 5x)$.
Show answer
This is a polynomial, so use direct substitution: \[ 4(3)^2 - 5(3) = 36 - 15 = 21. \]
Boxed answer: $21$. (This is Stewart 1.6, Exercise 3.)
Problem 2. Evaluate $\displaystyle\lim_{x\to 3}\frac{x - 3}{x - 3}$.
Show answer
For $x \neq 3$, the expression equals $1$. The limit ignores the single point $x = 3$, so \[ \lim_{x\to 3}\frac{x - 3}{x - 3} = \lim_{x\to 3} 1 = 1. \]
Boxed answer: $1$. (The function is undefined at $x = 3$, but the limit is still $1$.)
Problem 3. Evaluate $\displaystyle\lim_{h\to 4}\frac{h^2 - 2h - 8}{h - 4}$.
Show answer
Substitution gives $\frac00$. Factor the numerator: $h^2 - 2h - 8 = (h - 4)(h + 2)$. \[ \frac{(h - 4)(h + 2)}{h - 4} = h + 2 \quad (h \neq 4). \] \[ \lim_{h\to 4}\frac{h^2 - 2h - 8}{h - 4} = \lim_{h\to 4}(h + 2) = 6. \]
Boxed answer: $6$. (This is Stewart 1.6, Exercise 13.)
Level 2 -- Multiple Steps
Problem 4. Evaluate $\displaystyle\lim_{x\to -3}\frac{x^2 + 3x}{x^2 - x - 12}$.
Show answer
Substitution gives $\frac00$. Factor both: \[ \frac{x^2 + 3x}{x^2 - x - 12} = \frac{x(x + 3)}{(x - 4)(x + 3)} = \frac{x}{x - 4} \quad (x \neq -3). \] Substitute $x = -3$: \[ \frac{-3}{-3 - 4} = \frac{-3}{-7} = \frac{3}{7}. \]
Boxed answer: $\dfrac{3}{7}$. (This is Stewart 1.6, Exercise 14.)
Problem 5. Evaluate $\displaystyle\lim_{h\to 0}\frac{\sqrt{9 + h} - 3}{h}$.
Show answer
Substitution gives $\frac00$. Multiply by the conjugate $\sqrt{9+h} + 3$: \[ \frac{\sqrt{9 + h} - 3}{h}\cdot\frac{\sqrt{9 + h} + 3}{\sqrt{9 + h} + 3} = \frac{(9 + h) - 9}{h(\sqrt{9 + h} + 3)} = \frac{h}{h(\sqrt{9 + h} + 3)} = \frac{1}{\sqrt{9 + h} + 3}. \] Substitute $h = 0$: \[ \lim_{h\to 0}\frac{\sqrt{9 + h} - 3}{h} = \frac{1}{\sqrt{9} + 3} = \frac{1}{6}. \]
Boxed answer: $\dfrac{1}{6}$. (This is Stewart 1.6, Exercise 23.)
Problem 6. Let $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 1 \\ (x - 2)^2 & \text{if } x \geq 1 \end{cases}$. Find $\lim_{x\to 1^-} f(x)$ and $\lim_{x\to 1^+} f(x)$, then state whether $\lim_{x\to 1} f(x)$ exists.
Show answer
From the left ($x < 1$), use $x^2 + 1$: \[ \lim_{x\to 1^-} f(x) = 1^2 + 1 = 2. \] From the right ($x \geq 1$), use $(x - 2)^2$: \[ \lim_{x\to 1^+} f(x) = (1 - 2)^2 = 1. \] The one-sided limits differ ($2 \neq 1$), so the two-sided limit does not exist.
Boxed answer: left limit $2$, right limit $1$; $\lim_{x\to 1} f(x)$ does not exist. (This is Stewart 1.6, Exercise 52.)
Level 3 -- Deeper Problems
Problem 7. Evaluate $\displaystyle\lim_{h\to 0}\frac{(x + h)^3 - x^3}{h}$ (treat $x$ as a constant).
Show answer
Expand $(x+h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3$: \[ \frac{(x + h)^3 - x^3}{h} = \frac{3x^2 h + 3x h^2 + h^3}{h} = \frac{h(3x^2 + 3xh + h^2)}{h} = 3x^2 + 3xh + h^2 \quad (h \neq 0). \] Substitute $h = 0$: \[ \lim_{h\to 0}\frac{(x + h)^3 - x^3}{h} = 3x^2. \]
Boxed answer: $3x^2$. (This is Stewart 1.6, Exercise 32. It is the derivative of $x^3$, found from the definition.)
Problem 8. Use the Squeeze Theorem to show that $\displaystyle\lim_{x\to 0} x^2 \cos(20\pi x) = 0$.
Show answer
Since $-1 \leq \cos(20\pi x) \leq 1$ for all $x$, and $x^2 \geq 0$, multiply through by $x^2$: \[ -x^2 \leq x^2\cos(20\pi x) \leq x^2. \] Both bounds tend to zero: $\lim_{x\to 0}(-x^2) = 0$ and $\lim_{x\to 0} x^2 = 0$. By the Squeeze Theorem, \[ \lim_{x\to 0} x^2\cos(20\pi x) = 0. \]
Boxed answer: $0$. (This is Stewart 1.6, Exercise 37.)
Problem 9. If $2x \leq g(x) \leq x^4 - x^2 + 2$ for all $x$, find $\displaystyle\lim_{x\to 1} g(x)$.
Show answer
Evaluate the limit of each bounding function at $x = 1$ (both are polynomials, so substitute): \[ \lim_{x\to 1} 2x = 2, \qquad \lim_{x\to 1}(x^4 - x^2 + 2) = 1 - 1 + 2 = 2. \] The two bounds share the same limit $2$ at the target, so the hypothesis of the Squeeze Theorem is met. Since $2x \leq g(x) \leq x^4 - x^2 + 2$ and both ends head to $2$, \[ \lim_{x\to 1} g(x) = 2. \]
Boxed answer: $2$. (This is Stewart 1.6, Exercise 40.) The key step is confirming the upper and lower bounds agree at $x = 1$ before invoking the Squeeze Theorem; if they had disagreed, the bounds alone would not determine the limit.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Think of the limit laws as arithmetic that survives the limiting process.
A limit asks where a function is heading. The laws say that if you know where two functions are heading, you know where their sum, difference, product, and quotient are heading: apply the arithmetic operation to the two destinations. Nothing surprising happens, as long as you never divide by something heading to zero.
Three things follow from the picture:
- For nice functions, the destination is the value. A polynomial or a rational function (at a point in its domain) heads exactly to its value there. That is the Direct Substitution Property, and it makes most limits a one-step substitution.
- $\frac00$ is a request for algebra, not an answer. When substitution collapses to $\frac00$, the function is hiding a removable factor or a root that has not been cleared. Rewriting it into an equal-everywhere-else function reveals the true destination.
- A limit never looks at the single target point. It only watches the approach. That is why canceling a factor that vanishes at $a$ is legal, why a function with a hole can still have a limit, and why the Squeeze Theorem can pin a value the function may never actually take.
Connections
Within Functions and Limits (Chapter 1)
- One-sided limits: A two-sided limit exists exactly when the left and right limits agree. That existence theorem is how you settle limits of absolute-value, piecewise, and step functions.
- Direct substitution and algebraic techniques: The Direct Substitution Property and the factor, rationalize, and squeeze repairs each have their own node; the laws those techniques rest on are stated here.
- Continuity: A function is continuous at $a$ exactly when $\lim_{x\to a} f(x) = f(a)$, which is the Direct Substitution Property promoted to a definition. The next sections build continuity directly on these laws.
Toward later calculus (MATH161 and beyond)
- The derivative: Every derivative is a limit of a difference quotient, which always starts as $\frac00$ and is repaired by canceling $h$ (Examples 3 and Problem 7). The limit laws are what make the definition of the derivative computable.
- Limits at infinity: The same laws hold as $x \to \pm\infty$ (with the same caution about $\frac{\infty}{\infty}$ and $\infty - \infty$), which is what lets you evaluate horizontal-asymptote limits by dividing by the highest power.
- Continuity of combinations: Because limits respect arithmetic, sums, products, quotients, and compositions of continuous functions are continuous. The entire toolkit of “this function is continuous, so I can substitute” descends from these seven laws.
Audience Notes
For students who find math intimidating: Most limits are easier than they look. First, just try plugging in; if you get a number, that is the answer (the Direct Substitution Property). You only need the harder techniques when plugging in gives $\frac00$, and then there are just three moves to try: factor and cancel, multiply by the conjugate, or squeeze. Match the move to what you see, a factorable polynomial, a root, or an oscillating piece.
For career-focused students: These laws are why symbolic-math software (and a careful engineer) can simplify an expression before evaluating it, rather than blindly substituting and hitting a divide-by-zero. The $\frac00$ repairs are the same algebra a computer algebra system performs to find a limit, and recognizing a removable discontinuity is a routine task in signal and control analysis.
For gifted and curious students: The Squeeze Theorem proves the foundational trigonometric limit $\lim_{x\to 0}\frac{\sin x}{x} = 1$ by trapping $\frac{\sin x}{x}$ between $\cos x$ and $1$. That limit, in turn, is what gives the derivative of $\sin x$. Trace the chain: Squeeze Theorem, then the sine limit, then the derivative of sine, and you have built a piece of calculus from these laws alone.
For PhD-track students: Each law is a theorem with an $\varepsilon$-$\delta$ proof. The sum law, for instance, follows from the triangle inequality by splitting a target $\varepsilon$ into two halves. Proving the product and quotient laws rigorously, and seeing exactly where the nonzero-denominator hypothesis is used, is the standard first exercise in a real-analysis course and shows why the informal “limits respect arithmetic” is genuinely true.
Back to Functions and Limits | Related: Algebraic Limit Techniques | Next: Direct Substitution