The Central Idea
Suppose
\[ \lim_{x\to a}f(x)=L \]and
\[ \lim_{x\to a}g(x)=M. \]Limit laws tell us how the limits of combinations of \(f\) and \(g\) are related to \(L\) and \(M\).
In many cases, the rule is simple: take the limit first, then perform the algebraic operation.
The Constant Law
The limit of a constant is the constant itself.
\[ \boxed{\lim_{x\to a}c=c}. \]For example,
\[ \lim_{x\to5}7=7. \]The Identity Law
The limit of the identity function is the target input.
\[ \boxed{\lim_{x\to a}x=a}. \]For example,
\[ \lim_{x\to3}x=3. \]The Sum Law
If the individual limits exist, then
\[ \boxed{ \lim_{x\to a}[f(x)+g(x)] = \lim_{x\to a}f(x) + \lim_{x\to a}g(x) }. \]Using \(L\) and \(M\),
\[ \boxed{\lim_{x\to a}[f(x)+g(x)]=L+M}. \]Example: Sum Law
Suppose
\[ \lim_{x\to2}f(x)=5 \]and
\[ \lim_{x\to2}g(x)=-1. \]Then
\[ \lim_{x\to2}[f(x)+g(x)] = 5+(-1) = \boxed{4}. \]The Difference Law
Limits distribute over subtraction:
\[ \boxed{ \lim_{x\to a}[f(x)-g(x)] = \lim_{x\to a}f(x) - \lim_{x\to a}g(x) }. \]Therefore,
\[ \boxed{\lim_{x\to a}[f(x)-g(x)]=L-M}. \]Example: Difference Law
If
\[ \lim_{x\to4}f(x)=9 \]and
\[ \lim_{x\to4}g(x)=3, \]then
\[ \lim_{x\to4}[f(x)-g(x)] = 9-3 = \boxed{6}. \]The Constant-Multiple Law
A constant factor can be pulled through a limit:
\[ \boxed{ \lim_{x\to a}[c\,f(x)] = c\lim_{x\to a}f(x) }. \]Therefore,
\[ \boxed{\lim_{x\to a}[c\,f(x)]=cL}. \]Example: Constant Multiple
If
\[ \lim_{x\to1}f(x)=7, \]then
\[ \lim_{x\to1}[-3f(x)] = -3(7) = \boxed{-21}. \]The Product Law
Limits distribute over multiplication:
\[ \boxed{ \lim_{x\to a}[f(x)g(x)] = \left(\lim_{x\to a}f(x)\right) \left(\lim_{x\to a}g(x)\right) }. \]Thus,
\[ \boxed{\lim_{x\to a}[f(x)g(x)]=LM}. \]Example: Product Law
Suppose
\[ \lim_{x\to3}f(x)=4 \]and
\[ \lim_{x\to3}g(x)=-2. \]Then
\[ \lim_{x\to3}[f(x)g(x)] = 4(-2) = \boxed{-8}. \]The Quotient Law
Limits distribute over division provided the denominator limit is not zero:
\[ \boxed{ \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)} {\lim_{x\to a}g(x)} }. \]The condition is
\[ \boxed{M\neq0}. \]Therefore,
\[ \boxed{ \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M} }. \]Why the Denominator Condition Matters
Division by zero is not defined.
If
\[ \lim_{x\to a}g(x)=0, \]the quotient law cannot be applied directly.
The limit might still exist, but additional analysis is required.
Example: Quotient Law
Suppose
\[ \lim_{x\to2}f(x)=10 \]and
\[ \lim_{x\to2}g(x)=5. \]Then
\[ \lim_{x\to2}\frac{f(x)}{g(x)} = \frac{10}{5} = \boxed{2}. \]The Power Law
For a positive integer \(n\),
\[ \boxed{ \lim_{x\to a}[f(x)]^n = \left(\lim_{x\to a}f(x)\right)^n }. \]Therefore,
\[ \boxed{\lim_{x\to a}[f(x)]^n=L^n}. \]Example: Power Law
If
\[ \lim_{x\to3}f(x)=2, \]then
\[ \lim_{x\to3}[f(x)]^4 = 2^4 = \boxed{16}. \]The Root Law
Under the appropriate real-domain conditions,
\[ \boxed{ \lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to a}f(x)} }. \]Thus,
\[ \boxed{ \lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{L} }. \]Even Roots Require Nonnegative Inputs
For an even root such as
\[ \sqrt{f(x)}, \]the real-valued function requires
\[ f(x)\ge0. \]The domain must therefore be respected when applying the root law.
Example: Root Law
Evaluate
\[ \lim_{x\to4}\sqrt{x+5}. \]Since
\[ \lim_{x\to4}(x+5)=9, \]we have
\[ \boxed{\sqrt{9}=3}. \]Absolute-Value Law
Since the absolute-value function is continuous,
\[ \boxed{ \lim_{x\to a}|f(x)| = \left| \lim_{x\to a}f(x) \right| }. \]Therefore,
\[ \boxed{\lim_{x\to a}|f(x)|=|L|}. \]Example: Absolute Value
If
\[ \lim_{x\to2}f(x)=-7, \]then
\[ \lim_{x\to2}|f(x)| = |-7| = \boxed{7}. \]Polynomial Limits
Because polynomials are built from constants, powers, sums, and products, the limit laws imply:
\[ \boxed{\lim_{x\to a}P(x)=P(a)}. \]This is the algebraic reason direct substitution works for every polynomial.
Example: Polynomial by Limit Laws
Evaluate
\[ \lim_{x\to2}(3x^2-4x+5). \]Apply the laws term by term:
\[ 3(2^2)-4(2)+5. \]Therefore,
\[ \boxed{9}. \]Rational Function Limits
If
\[ R(x)=\frac{P(x)}{Q(x)} \]and
\[ Q(a)\neq0, \]then
\[ \boxed{ \lim_{x\to a}R(x) = \frac{P(a)}{Q(a)} }. \]Example: Rational Function
Evaluate
\[ \lim_{x\to2} \frac{x^2+3}{x+1}. \]The denominator approaches
\[ 3\neq0. \]Therefore,
\[ \frac{2^2+3}{2+1} = \frac{7}{3}. \]So
\[ \boxed{\frac73}. \]When Direct Substitution Fails
Limit laws do not eliminate the need for algebraic analysis.
Consider
\[ \lim_{x\to2} \frac{x^2-4}{x-2}. \]Direct substitution gives
\[ \frac00. \]The quotient law cannot be applied directly because the denominator limit is zero.
Use Algebra First, Then Apply the Laws
Factor:
\[ \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2}. \]For \(x\neq2\),
\[ \frac{x^2-4}{x-2}=x+2. \]Now the limit laws apply:
\[ \lim_{x\to2}(x+2)=4. \]Therefore,
\[ \boxed{4}. \]Limit Laws and One-Sided Limits
The same algebraic laws work for one-sided limits whenever the required one-sided limits exist.
For example, if
\[ \lim_{x\to a^+}f(x)=L \]and
\[ \lim_{x\to a^+}g(x)=M, \]then
\[ \lim_{x\to a^+}[f(x)+g(x)] = L+M. \]Limit Laws at Infinity
The same structural rules also apply to limits as
\[ x\to+\infty \]or
\[ x\to-\infty, \]provided the individual limits and operations are well-defined.
Composition Law
Suppose
\[ \lim_{x\to a}g(x)=L \]and \(f\) is continuous at \(L\).
Then
\[ \boxed{ \lim_{x\to a}f(g(x)) = f\left(\lim_{x\to a}g(x)\right) = f(L) }. \]Example: Composition
Evaluate
\[ \lim_{x\to2}\sqrt{x^2+5}. \]First,
\[ \lim_{x\to2}(x^2+5)=9. \]Since the square-root function is continuous at \(9\),
\[ \boxed{\sqrt{9}=3}. \]Nested Functions Can Be Handled Layer by Layer
Consider
\[ \lim_{x\to1} \left(2x^2+3\right)^4. \]First evaluate the inner polynomial:
\[ 2(1)^2+3=5. \]Then apply the power:
\[ 5^4=625. \]Therefore,
\[ \boxed{625}. \]Repeated Use of the Laws
Complicated expressions are usually evaluated by applying several limit laws in sequence.
For example,
\[ \lim_{x\to2} \frac{3x^2+4x-1}{x+5} \]uses:
- the power law;
- the constant-multiple law;
- the sum and difference laws;
- the quotient law.
Example: Combined Laws
Evaluate
\[ \lim_{x\to2} \frac{3x^2+4x-1}{x+5}. \]The numerator limit is
\[ 3(4)+8-1=19. \]The denominator limit is
\[ 7. \]Therefore,
\[ \boxed{\frac{19}{7}}. \]Limit Laws Explain Direct Substitution
Direct substitution is not a separate mysterious trick.
It works because many standard functions are built from operations that preserve limits.
When the expression remains defined at the target, the limit laws often reduce evaluation to ordinary arithmetic.
When the Laws Do Not Apply Automatically
The rules have conditions.
For example:
- the quotient law requires a nonzero denominator limit;
- even-root limits require appropriate real-domain conditions;
- individual component limits must exist when a law assumes them;
- indeterminate forms require additional analysis.
Limit Laws Do Not Create Missing Limits
Suppose
\[ \lim_{x\to a}f(x) \]does not exist.
We cannot automatically use a sum, product, or quotient law involving that missing limit.
Sometimes the combined expression may still have a limit, but it must be analyzed directly.
Example: Components May Fail While the Combination Works
Consider
\[ f(x)=\frac{1}{x} \]and
\[ g(x)=-\frac{1}{x} \]near \(x=0\).
Neither function has a finite two-sided limit at \(0\).
But
\[ f(x)+g(x)=0 \]for every \(x\neq0\).
Therefore,
\[ \lim_{x\to0}[f(x)+g(x)]=0. \]Why This Does Not Violate the Sum Law
The sum law says that if both individual limits exist, then the limit of the sum equals the sum of the limits.
It does not say that the sum cannot have a limit when the individual limits fail to exist.
The law gives a sufficient calculation rule, not a universal reverse test.
Known Limits Can Be Treated as Building Blocks
If
\[ \lim_{x\to a}f(x)=L \]and
\[ \lim_{x\to a}g(x)=M, \]then we can immediately evaluate expressions such as
\[ 3f(x)-2g(x), \] \[ [f(x)]^2g(x), \]or
\[ \frac{f(x)+1}{g(x)} \]when the required conditions hold.
Example from Known Limits
Suppose
\[ \lim_{x\to5}f(x)=3 \]and
\[ \lim_{x\to5}g(x)=4. \]Evaluate
\[ \lim_{x\to5}[2f(x)-3g(x)]. \]Using the constant-multiple and difference laws:
\[ 2(3)-3(4)=6-12=-6. \]Therefore,
\[ \boxed{-6}. \]Example with Product and Power
Using the same limits,
\[ \lim_{x\to5}[f(x)]^2g(x) = (3)^2(4) = \boxed{36}. \]Example with a Quotient
Again suppose
\[ \lim_{x\to5}f(x)=3 \]and
\[ \lim_{x\to5}g(x)=4. \]Then
\[ \lim_{x\to5} \frac{f(x)+1}{g(x)} = \frac{3+1}{4} = \boxed{1}. \]A Reliable Procedure
- Identify the outermost algebraic operation.
- Break the expression into simpler component limits.
- Apply the appropriate limit laws.
- Check any required conditions, especially denominator restrictions.
- Use direct substitution when the laws reduce the expression to ordinary arithmetic.
- If an indeterminate form appears, simplify before continuing.
- Reassemble the component limits carefully.
Example: Full Limit-Law Workflow
Evaluate
\[ \lim_{x\to3} \frac{2x^2-5x+4}{x+2}. \]The numerator limit is
\[ 2(3^2)-5(3)+4 = 18-15+4 = 7. \]The denominator limit is
\[ 3+2=5. \]Since the denominator limit is nonzero,
\[ \boxed{\frac75}. \]Common Error: Forgetting the Quotient Condition
The quotient law cannot be applied directly when the denominator limit is zero.
In
\[ \lim_{x\to2}\frac{x^2-4}{x-2}, \]direct quotient-law substitution gives \(0/0\), which is indeterminate.
Simplify first.
Common Error: Applying a Law to a Missing Component Limit
If one of the required individual limits does not exist, the corresponding limit law cannot simply be invoked.
Analyze the combined expression directly instead.
Common Error: Losing Parentheses
When substituting limit values, preserve the original operation structure.
For example,
\[ [f(x)+g(x)]^2 \]becomes
\[ (L+M)^2, \]not
\[ L+M^2. \]Common Error: Ignoring Domain Restrictions
The root law and quotient law depend on the expression being meaningful in the relevant real domain.
Always check whether the original function is defined near the target from the required direction.
Common Error: Treating the Laws as Mere Memorization
The laws are not isolated formulas.
They express a deeper principle: standard algebraic operations preserve limiting behavior under appropriate conditions.
Understanding that structure makes the rules easier to use and remember.
Practice 1: Constant and Identity Laws
Evaluate:
\[ \lim_{x\to7}(3x+5). \]Practice 2: Polynomial Limit
Evaluate:
\[ \lim_{x\to2}(x^3-4x+1). \]Practice 3: Quotient Law
Evaluate:
\[ \lim_{x\to3} \frac{x^2+1}{x+4}. \]Practice 4: Known Limits
Suppose
\[ \lim_{x\to1}f(x)=2 \]and
\[ \lim_{x\to1}g(x)=5. \]Evaluate
\[ \lim_{x\to1}[3f(x)+2g(x)]. \]Practice 5: Product and Power
Using the same limits, evaluate
\[ \lim_{x\to1}[f(x)]^2g(x). \]Practice 6: Quotient from Known Limits
Using the same limits, evaluate
\[ \lim_{x\to1} \frac{g(x)-1}{f(x)}. \]Practice 7: Root Law
Evaluate:
\[ \lim_{x\to4}\sqrt{2x+1}. \]Practice 8: Absolute Value
Suppose
\[ \lim_{x\to2}h(x)=-6. \]Evaluate
\[ \lim_{x\to2}|h(x)|. \]Practice 9: Identify the Problem
Explain why the quotient law cannot be applied directly to
\[ \lim_{x\to5}\frac{x^2-25}{x-5}. \]Practice 10: Full Workflow
Evaluate:
\[ \lim_{x\to2} \frac{4x^2-3x+1}{x^2+2}. \]Practice Answers
Practice 1
\[ 3(7)+5=26. \]Therefore,
\[ \boxed{26}. \]Practice 2
\[ 2^3-4(2)+1 = 8-8+1 = \boxed{1}. \]Practice 3
\[ \frac{3^2+1}{3+4} = \frac{10}{7}. \]Therefore,
\[ \boxed{\frac{10}{7}}. \]Practice 4
\[ 3(2)+2(5) = 6+10 = \boxed{16}. \]Practice 5
\[ (2)^2(5) = 4\cdot5 = \boxed{20}. \]Practice 6
\[ \frac{5-1}{2} = \boxed{2}. \]Practice 7
\[ \sqrt{2(4)+1} = \sqrt{9} = \boxed{3}. \]Practice 8
\[ |-6|=\boxed{6}. \]Practice 9
Direct substitution makes the denominator limit equal to zero:
\[ \frac{0}{0}. \]The quotient law requires a nonzero denominator limit, so the expression must first be simplified.
Practice 10
The numerator limit is
\[ 4(2^2)-3(2)+1 = 16-6+1 = 11. \]The denominator limit is
\[ 2^2+2=6. \]Therefore,
\[ \boxed{\frac{11}{6}}. \]What to Remember
If
\[ \lim_{x\to a}f(x)=L \qquad\text{and}\qquad \lim_{x\to a}g(x)=M, \]then the main laws include:
\[ \boxed{ \lim(f+g)=L+M } \] \[ \boxed{ \lim(f-g)=L-M } \] \[ \boxed{ \lim(cf)=cL } \] \[ \boxed{ \lim(fg)=LM } \] \[ \boxed{ \lim\left(\frac{f}{g}\right)=\frac{L}{M}, \quad M\neq0 } \] \[ \boxed{ \lim[f(x)]^n=L^n }. \]- Limit laws let complicated limits be assembled from simpler ones.
- They explain why direct substitution works for continuous algebraic expressions.
- The quotient law requires a nonzero denominator limit.
- Domain restrictions still matter for roots and other functions.
- Indeterminate forms require simplification before the laws can finish the problem.
The central strategy is: decompose the expression, apply the appropriate laws, check the conditions, and then recombine the results.
Where We Go Next
Limit laws handle many expressions directly, but some limits are difficult to evaluate even after ordinary algebraic simplification. In those cases, comparison can provide the answer.
The next lesson, The Squeeze Theorem, shows how a difficult function can be trapped between two simpler functions that approach the same limit.