The Central Idea
To understand what happens near \(x=a\), we can approach \(a\) from two separate directions.
From the left:
\[ \boxed{\lim_{x\to a^-}f(x)} \]From the right:
\[ \boxed{\lim_{x\to a^+}f(x)}. \]A finite two-sided limit exists only when the left-hand and right-hand limits agree.
\[ \boxed{ \lim_{x\to a}f(x)=L \quad\text{when}\quad \lim_{x\to a^-}f(x)= \lim_{x\to a^+}f(x)=L }. \]What the Minus and Plus Signs Mean
In
\[ x\to a^-, \]the superscript minus means that \(x\) approaches \(a\) through values less than \(a\).
In
\[ x\to a^+, \]the superscript plus means that \(x\) approaches \(a\) through values greater than \(a\).
These signs describe the direction of approach. They do not mean that \(a\) itself is negative or positive.
Approaching from the Left
To estimate
\[ \lim_{x\to3^-}f(x), \]we might use inputs such as
\[ 2.9,\quad2.99,\quad2.999. \]Every input is less than \(3\), but the values move progressively closer to \(3\).
Approaching from the Right
To estimate
\[ \lim_{x\to3^+}f(x), \]we might use
\[ 3.1,\quad3.01,\quad3.001. \]Every input is greater than \(3\), but again the values move progressively closer to \(3\).
A Simple One-Sided Example
Consider
\[ f(x)=x^2. \]As \(x\) approaches \(2\) from the left,
\[ f(x)\to4. \]As \(x\) approaches \(2\) from the right,
\[ f(x)\to4. \]Therefore,
\[ \lim_{x\to2^-}x^2=4, \qquad \lim_{x\to2^+}x^2=4, \]and hence
\[ \boxed{\lim_{x\to2}x^2=4}. \]When the Two Sides Disagree
Suppose
\[ \lim_{x\to4^-}f(x)=2 \]while
\[ \lim_{x\to4^+}f(x)=7. \]The function approaches different values from the two directions.
Therefore,
\[ \boxed{\lim_{x\to4}f(x)\text{ does not exist}}. \]The Two-Sided Limit Test
For a finite limit, the rule is:
\[ \boxed{ \lim_{x\to a}f(x)=L \iff \lim_{x\to a^-}f(x)=L \text{ and } \lim_{x\to a^+}f(x)=L }. \]The symbol \(\iff\) means “if and only if.”
This criterion is one of the most useful tests in elementary limit analysis.
A Piecewise Example
Suppose
\[ f(x)= \begin{cases} x+1, & x<2,\\ 5-x, & x\ge2. \end{cases} \]For the left-hand limit, use the formula valid when \(x<2\):
\[ \lim_{x\to2^-}f(x) = \lim_{x\to2^-}(x+1) = 3. \]For the right-hand limit, use the formula valid when \(x\ge2\):
\[ \lim_{x\to2^+}f(x) = \lim_{x\to2^+}(5-x) = 3. \]Since the two sides agree,
\[ \boxed{\lim_{x\to2}f(x)=3}. \]A Piecewise Jump
Now consider
\[ g(x)= \begin{cases} x+1, & x<2,\\ x+4, & x\ge2. \end{cases} \]From the left,
\[ \lim_{x\to2^-}g(x)=3. \]From the right,
\[ \lim_{x\to2^+}g(x)=6. \]Therefore,
\[ \boxed{\lim_{x\to2}g(x)\text{ does not exist}}. \]This is the basic pattern of a jump discontinuity.
Graphical One-Sided Limits
On a graph, a left-hand limit is found by tracing the graph toward the target \(x\)-value from smaller \(x\)-values.
A right-hand limit is found by tracing toward the same point from larger \(x\)-values.
The actual filled point at the target still does not determine either one-sided limit by itself.
Endpoints Naturally Use One-Sided Limits
Suppose a function is defined only for
\[ x\ge0. \]At the endpoint \(x=0\), there are no domain values immediately to the left.
The relevant local behavior is therefore
\[ \lim_{x\to0^+}f(x). \]One-sided limits are especially important at endpoints of restricted domains.
Example with a Square Root
Consider
\[ f(x)=\sqrt{x}. \]Over the real numbers, the domain is
\[ x\ge0. \]Thus,
\[ \boxed{\lim_{x\to0^+}\sqrt{x}=0}. \]A real left-hand limit at \(0\) is not available because the function is not defined for negative real inputs near \(0\).
Infinite Limits
Sometimes function values do not approach a finite number. Instead, their magnitude grows without bound.
We write
\[ \boxed{\lim_{x\to a}f(x)=+\infty} \]when \(f(x)\) becomes arbitrarily large and positive as \(x\) approaches \(a\).
Similarly,
\[ \boxed{\lim_{x\to a}f(x)=-\infty} \]means the outputs decrease without bound.
Infinity Is Not an Ordinary Real Number
The symbols
\[ +\infty \qquad\text{and}\qquad -\infty \]describe unbounded behavior.
They are not ordinary real-number outputs that a function simply reaches.
An infinite limit tells us that the function can exceed every positive bound, or fall below every negative bound, sufficiently close to the target.
The Basic Reciprocal Model
Consider
\[ f(x)=\frac{1}{x}. \]As \(x\) approaches \(0\) from the right, \(x\) is a small positive number.
Its reciprocal becomes very large and positive:
\[ \frac{1}{0.1}=10, \qquad \frac{1}{0.01}=100, \qquad \frac{1}{0.001}=1000. \]Therefore,
\[ \boxed{\lim_{x\to0^+}\frac{1}{x}=+\infty}. \]The Reciprocal from the Left
When \(x\) approaches \(0\) from the left, \(x\) is a small negative number.
\[ \frac{1}{-0.1}=-10, \qquad \frac{1}{-0.01}=-100, \qquad \frac{1}{-0.001}=-1000. \]Therefore,
\[ \boxed{\lim_{x\to0^-}\frac{1}{x}=-\infty}. \]Opposite Infinite Behavior
For
\[ f(x)=\frac{1}{x}, \]we have
\[ \lim_{x\to0^-}\frac{1}{x}=-\infty \]and
\[ \lim_{x\to0^+}\frac{1}{x}=+\infty. \]The two sides do not agree.
Thus there is no single two-sided limit at \(x=0\).
Same-Sign Infinite Behavior
Consider
\[ f(x)=\frac{1}{x^2}. \]Whether \(x\) approaches \(0\) from the left or the right,
\[ x^2 \]is a small positive number.
Therefore,
\[ \lim_{x\to0^-}\frac{1}{x^2}=+\infty \]and
\[ \lim_{x\to0^+}\frac{1}{x^2}=+\infty. \]We summarize this behavior as
\[ \boxed{\lim_{x\to0}\frac{1}{x^2}=+\infty}. \]Vertical Asymptotes
If a function grows without bound as \(x\) approaches \(a\) from at least one side, the vertical line
\[ \boxed{x=a} \]is a vertical asymptote of the graph.
For
\[ f(x)=\frac{1}{x}, \]the line
\[ x=0 \]is a vertical asymptote.
Vertical Asymptotes and Denominators
Rational functions often have vertical asymptotes where a denominator approaches zero while the numerator remains nonzero.
For example,
\[ f(x)=\frac{3}{x-2} \]has a denominator approaching zero when
\[ x\to2. \]The behavior on each side depends on the sign of \(x-2\).
Analyzing \(\frac{3}{x-2}\) from the Left
If \(x<2\) but close to \(2\), then
\[ x-2 \]is a small negative number.
Therefore,
\[ \frac{3}{x-2} \]is a large negative number.
\[ \boxed{\lim_{x\to2^-}\frac{3}{x-2}=-\infty}. \]Analyzing \(\frac{3}{x-2}\) from the Right
If \(x>2\) but close to \(2\), then
\[ x-2 \]is a small positive number.
Therefore,
\[ \boxed{\lim_{x\to2^+}\frac{3}{x-2}=+\infty}. \]Sign Analysis Is Often the Fastest Method
For an expression such as
\[ \frac{5}{x+4}, \]the denominator changes sign at
\[ x=-4. \]Immediately to the left of \(-4\), \(x+4<0\).
Immediately to the right, \(x+4>0\).
Since the numerator \(5\) is positive,
\[ \lim_{x\to-4^-}\frac{5}{x+4}=-\infty, \qquad \lim_{x\to-4^+}\frac{5}{x+4}=+\infty. \]Squared Denominators Behave Differently
Consider
\[ \frac{5}{(x+4)^2}. \]The squared denominator is positive on both sides of \(-4\).
As \(x\to-4\), the denominator approaches zero through positive values.
Therefore,
\[ \boxed{ \lim_{x\to-4^-}\frac{5}{(x+4)^2} = \lim_{x\to-4^+}\frac{5}{(x+4)^2} = +\infty }. \]Odd and Even Powers Near a Vertical Asymptote
The parity of a factor can help predict sign behavior.
For
\[ \frac{1}{(x-a)^n}, \]- if \(n\) is even, the denominator has the same positive sign on both sides;
- if \(n\) is odd, the denominator changes sign across \(x=a\).
This provides a quick structural shortcut for many rational limits.
A Numerator Can Reverse the Sign
Consider
\[ f(x)=\frac{-2}{(x-1)^2}. \]The denominator is positive and approaches \(0\) from above on both sides.
The numerator is negative.
Therefore,
\[ \boxed{ \lim_{x\to1^-}f(x) = \lim_{x\to1^+}f(x) = -\infty }. \]Not Every Zero Denominator Creates a Vertical Asymptote
Consider
\[ \frac{x^2-4}{x-2}. \]Both numerator and denominator vanish at \(x=2\).
Factoring gives
\[ \frac{(x-2)(x+2)}{x-2}=x+2 \qquad(x\neq2). \]Thus,
\[ \lim_{x\to2}\frac{x^2-4}{x-2}=4. \]The graph has a removable hole, not a vertical asymptote.
Hole versus Vertical Asymptote
A useful first distinction is:
- a factor that cancels completely often produces a removable hole;
- a denominator factor that remains after simplification can produce a vertical asymptote.
Algebraic simplification should therefore come before declaring a vertical asymptote.
Example: Simplify Before Classifying
Consider
\[ f(x)=\frac{(x-3)(x+1)}{(x-3)(x-5)}. \]For \(x\neq3\),
\[ f(x)=\frac{x+1}{x-5}. \]The factor \(x-3\) cancels, so \(x=3\) corresponds to a removable hole.
The factor \(x-5\) remains in the denominator, so \(x=5\) is a vertical asymptote.
One-Sided Limits at the Vertical Asymptote
For
\[ f(x)=\frac{x+1}{x-5}, \]near \(x=5\), the numerator is close to \(6\), which is positive.
The denominator is negative immediately to the left and positive immediately to the right.
Therefore,
\[ \boxed{ \lim_{x\to5^-}\frac{x+1}{x-5}=-\infty } \]and
\[ \boxed{ \lim_{x\to5^+}\frac{x+1}{x-5}=+\infty }. \]Numerical Evidence for an Infinite Limit
Consider
\[ f(x)=\frac{1}{(x-2)^2}. \]| \(x\) | \(f(x)\) |
|---|---|
| \(1.9\) | \(100\) |
| \(1.99\) | \(10{,}000\) |
| \(1.999\) | \(1{,}000{,}000\) |
| \(2.001\) | \(1{,}000{,}000\) |
| \(2.01\) | \(10{,}000\) |
| \(2.1\) | \(100\) |
The outputs grow rapidly without bound from both sides.
Graphical Appearance of Infinite Limits
Near a vertical asymptote, the graph typically rises or falls sharply as it approaches the asymptote.
A branch may rise toward
\[ +\infty \]or fall toward
\[ -\infty. \]The direction can differ on the two sides.
Infinite Limits Are Still Local
The statement
\[ \lim_{x\to2^+}f(x)=+\infty \]concerns values of \(f(x)\) for \(x\) close to \(2\) on the right.
It says nothing by itself about what happens far from \(x=2\).
Infinite limits at finite inputs are therefore still local descriptions.
Do Not Confuse Infinite Limits with Limits at Infinity
These two ideas use the word “infinity” differently.
An infinite limit has a finite input target but unbounded output:
\[ \lim_{x\to2}f(x)=+\infty. \]A limit at infinity asks what happens as the input itself grows without bound:
\[ \lim_{x\to+\infty}f(x). \]The next lesson develops the second idea.
A Reliable Procedure for One-Sided Limits
- Identify the target input \(a\).
- Determine whether the approach is from the left or right.
- Use only values from the requested side.
- For piecewise functions, select the formula valid on that side.
- Evaluate the behavior as \(x\) moves toward \(a\).
- Keep the one-sided answer separate from the two-sided conclusion.
A Reliable Procedure for Infinite Limits
- Check whether direct substitution produces a zero denominator.
- Simplify first to identify removable factors.
- Locate any denominator factor that remains zero at the target.
- Determine the sign of the numerator near the target.
- Determine the sign of the denominator on each side.
- Combine the signs to decide between \(+\infty\) and \(-\infty\).
- Confirm with a table or graph when useful.
Example: Full Sign-Analysis Workflow
Evaluate the one-sided limits of
\[ f(x)=\frac{x+2}{x-1} \]at \(x=1\).
Near \(x=1\), the numerator \(x+2\) is close to \(3\), so it is positive.
From the left,
\[ x-1<0, \]so
\[ \boxed{\lim_{x\to1^-}\frac{x+2}{x-1}=-\infty}. \]From the right,
\[ x-1>0, \]so
\[ \boxed{\lim_{x\to1^+}\frac{x+2}{x-1}=+\infty}. \]Example: Even-Power Denominator
Evaluate
\[ \lim_{x\to3^\pm}\frac{4}{(x-3)^2}. \]Because
\[ (x-3)^2>0 \]on both sides of \(3\), and the denominator approaches \(0\),
\[ \boxed{ \lim_{x\to3^-}\frac{4}{(x-3)^2}=+\infty } \]and
\[ \boxed{ \lim_{x\to3^+}\frac{4}{(x-3)^2}=+\infty }. \]Common Error: Treating \(a^-\) as a Negative Number
In
\[ x\to5^-, \]the minus sign does not mean \(-5\).
It means approaching \(5\) through values less than \(5\).
Common Error: Ignoring the Other Side
Knowing
\[ \lim_{x\to a^-}f(x)=L \]does not by itself establish
\[ \lim_{x\to a}f(x)=L. \]The right-hand limit must also exist and equal \(L\).
Common Error: Calling Every Undefined Point a Vertical Asymptote
An undefined point may be a removable hole instead.
Always simplify a rational expression before classifying the discontinuity.
A canceled factor can create a hole; a remaining denominator factor can create a vertical asymptote.
Common Error: Saying Infinity Is a Function Value
When we write
\[ \lim_{x\to a}f(x)=+\infty, \]we are describing unbounded growth.
We are not saying that \(f(a)\) equals an ordinary real number called infinity.
Common Error: Missing the Sign
The expressions
\[ +\infty \qquad\text{and}\qquad -\infty \]describe different directions of unbounded behavior.
Sign analysis is therefore essential near vertical asymptotes.
Practice 1: Left-Hand Limit
Suppose values of \(f(x)\) approach \(4\) as \(x\) approaches \(2\) through values less than \(2\).
Write the limit using proper notation.
Practice 2: Piecewise Function
Let
\[ f(x)= \begin{cases} 2x+1, & x<3,\\ x+4, & x\ge3. \end{cases} \]Find
\[ \lim_{x\to3^-}f(x) \]and
\[ \lim_{x\to3^+}f(x). \]Does the two-sided limit exist?
Practice 3: Jump Behavior
Suppose
\[ \lim_{x\to5^-}g(x)=1 \]and
\[ \lim_{x\to5^+}g(x)=6. \]Find
\[ \lim_{x\to5}g(x). \]Practice 4: Basic Infinite Limit
Evaluate:
\[ \lim_{x\to0^+}\frac{1}{x}. \]Practice 5: Opposite Side
Evaluate:
\[ \lim_{x\to0^-}\frac{1}{x}. \]Practice 6: Squared Denominator
Evaluate both one-sided limits:
\[ \lim_{x\to2^-}\frac{3}{(x-2)^2}, \qquad \lim_{x\to2^+}\frac{3}{(x-2)^2}. \]Practice 7: Sign Analysis
Evaluate:
\[ \lim_{x\to4^-}\frac{2}{x-4} \]and
\[ \lim_{x\to4^+}\frac{2}{x-4}. \]Practice 8: Hole or Vertical Asymptote?
Classify the behavior at \(x=3\):
\[ f(x)=\frac{x^2-9}{x-3}. \]Practice 9: Simplify Before Classifying
Consider
\[ g(x)=\frac{(x-2)(x+5)}{(x-2)(x+1)}. \]Identify the removable hole and the vertical asymptote.
Practice 10: Full One-Sided Analysis
Evaluate:
\[ \lim_{x\to-2^-}\frac{x+1}{x+2} \]and
\[ \lim_{x\to-2^+}\frac{x+1}{x+2}. \]Practice Answers
Practice 1
\[ \boxed{\lim_{x\to2^-}f(x)=4}. \]Practice 2
From the left, use \(2x+1\):
\[ \lim_{x\to3^-}f(x)=2(3)+1=7. \]From the right, use \(x+4\):
\[ \lim_{x\to3^+}f(x)=3+4=7. \]Since the sides agree,
\[ \boxed{\lim_{x\to3}f(x)=7}. \]Practice 3
The one-sided limits are different, so
\[ \boxed{\lim_{x\to5}g(x)\text{ does not exist}}. \]Practice 4
\[ \boxed{\lim_{x\to0^+}\frac{1}{x}=+\infty}. \]Practice 5
\[ \boxed{\lim_{x\to0^-}\frac{1}{x}=-\infty}. \]Practice 6
The squared denominator is positive on both sides:
\[ \boxed{ \lim_{x\to2^-}\frac{3}{(x-2)^2}=+\infty } \]and
\[ \boxed{ \lim_{x\to2^+}\frac{3}{(x-2)^2}=+\infty }. \]Practice 7
To the left of \(4\), \(x-4\) is a small negative number:
\[ \boxed{\lim_{x\to4^-}\frac{2}{x-4}=-\infty}. \]To the right of \(4\), \(x-4\) is a small positive number:
\[ \boxed{\lim_{x\to4^+}\frac{2}{x-4}=+\infty}. \]Practice 8
\[ x^2-9=(x-3)(x+3). \]The factor \(x-3\) cancels, so \(x=3\) is a removable hole rather than a vertical asymptote.
The limit is
\[ \boxed{6}. \]Practice 9
Cancel the common factor \(x-2\):
\[ g(x)=\frac{x+5}{x+1} \qquad(x\neq2). \]Therefore:
- \(x=2\) is a removable hole;
- \(x=-1\) is a vertical asymptote.
Practice 10
Near \(x=-2\), the numerator \(x+1\) is close to \(-1\), so it is negative.
From the left, \(x+2<0\). Negative divided by negative is positive:
\[ \boxed{\lim_{x\to-2^-}\frac{x+1}{x+2}=+\infty}. \]From the right, \(x+2>0\). Negative divided by positive is negative:
\[ \boxed{\lim_{x\to-2^+}\frac{x+1}{x+2}=-\infty}. \]What to Remember
One-sided limits separate the two directions of approach:
\[ \boxed{\lim_{x\to a^-}f(x)} \qquad\text{and}\qquad \boxed{\lim_{x\to a^+}f(x)}. \]- A finite two-sided limit exists only when both one-sided limits agree.
- Piecewise functions often require separate left-hand and right-hand analysis.
- Infinite limits describe unbounded output near a finite input.
- \(+\infty\) and \(-\infty\) describe different directions of unbounded behavior.
- Vertical asymptotes are associated with unbounded behavior near a finite input.
- Always simplify rational expressions before declaring a vertical asymptote.
- Sign analysis is one of the fastest tools for determining one-sided infinite behavior.
The most important checkpoint is: analyze the left and right sides separately before combining them into a two-sided conclusion.
Where We Go Next
Infinite limits describe what happens when the output grows without bound near a finite input. A different question asks what a function does when the input itself becomes arbitrarily large in magnitude.
The next lesson, Limits at Infinity, studies end behavior, horizontal asymptotes, and the long-run behavior of functions as \(x\to+\infty\) or \(x\to-\infty\).