Unit 1 Limits: How Calculus Describes Approaching, Not Just Plugging In

Defining Limits and Using Limit Notation

What a limit is (and what it is not)

A limit describes the value a function’s outputs approach as the input gets close to some number. The key idea is that you are allowed to care about what happens near a point without requiring the function to be well-behaved at the point.

For example, a function might have a hole at $x = 2$, or it might not even be defined there. You can still ask: “As $x$ gets close to $2$, what do the $y$-values get close to?” That “get close to” value (if it exists) is the limit.

This matters because calculus is built on “approaching” ideas:

• Derivatives are limits of average rates of change as the interval shrinks.
• Definite integrals are limits of Riemann sums as rectangles get thinner.
• Many functions are analyzed by their behavior near points where direct substitution fails.

A common misconception is to think a limit is just “plug in the number.” Direct substitution is a useful method when it works, but the definition of a limit is about behavior near the point.

Limit notation and language

The standard notation

$\lim_{x \to a} f(x) = L$

is read: “The limit of $f(x)$ as $x$ approaches $a$ equals $L$.”

Interpretation:

• $x \to a$ means $x$ gets close to $a$ (from either side unless stated otherwise).
• $f(x)$ is the output.
• $L$ is the value the outputs approach.

It helps to separate three ideas:

1. The input is near $a$ (not necessarily equal).
2. The outputs are near $L$.
3. The claim is that you can make outputs as close to $L$ as you want by taking inputs sufficiently close to $a$.

One-sided limits: approaching from left vs right

Sometimes the behavior differs depending on whether you approach from smaller or larger $x$-values. Then you use one-sided limits:

$\lim_{x \to a^-} f(x) = L$

means approaching $a$ from the left (values less than $a$).

$\lim_{x \to a^+} f(x) = M$

means approaching $a$ from the right (values greater than $a$).

A two-sided limit exists exactly when both one-sided limits exist and are equal:

$\lim_{x \to a} f(x)$ exists if and only if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$

If the left and right sides approach different numbers, the two-sided limit does not exist.

A quick notation reference table

Limits vs function values

It is completely possible that

• $\lim_{x \to a} f(x) = L$ exists,
• but $f(a)$ is different from $L$,
• or $f(a)$ doesn’t exist.

That’s not a contradiction, because the limit ignores what happens at exactly $x = a$.

Example 1: A removable discontinuity (a hole)

Consider

$f(x) = \frac{x^2 - 1}{x - 1}$

You cannot plug in $x = 1$ because the denominator becomes $0$, so $f(1)$ is undefined. But for $x \ne 1$ you can simplify:

$f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1$

So as $x$ approaches $1$, $f(x)$ behaves like $x + 1$, which approaches $2$.

$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$

Important takeaway: the limit exists even though $f(1)$ does not.

Example 2: When left and right disagree

Define a function (conceptually) where

• for $x < 0$, $f(x) = -1$
• for $x > 0$, $f(x) = 1$

Then

$\lim_{x \to 0^-} f(x) = -1$

and

$\lim_{x \to 0^+} f(x) = 1$

Since these are not equal, the two-sided limit does not exist:

$\lim_{x \to 0} f(x)$ does not exist.

Exam Focus

• Typical question patterns:
  • You’re asked to interpret a statement like $\lim_{x \to 3} f(x) = 5$ in words, or match it to a graph.
  • You’re given left- and right-hand limits and asked whether $\lim_{x \to a} f(x)$ exists.
  • You’re asked about the relationship between $\lim_{x \to a} f(x)$ and $f(a)$.
• Common mistakes:
  • Assuming the limit equals the function value automatically (forgetting holes or jumps).
  • Forgetting that a two-sided limit requires agreement from both sides.
  • Saying “the limit is undefined because the function is undefined at the point” (limits can exist without the function being defined there).

Estimating Limit Values from Graphs and Tables

Why estimation matters

In many problems, you don’t start with a nice algebraic formula. You might get a graph, a table of values, or even a real-world measurement situation. Limits are about trends, so these representations can be enough to determine (or approximate) a limit.

On the AP exam, graph/table limit questions test whether you understand the meaning of approaching and whether you can distinguish:

• approaching a value vs actually reaching it,
• left-hand vs right-hand behavior,
• limits that don’t exist vs limits that are infinite or oscillatory.

Reading limits from a graph

To estimate $\lim_{x \to a} f(x)$ from a graph:

1. Locate $x = a$ on the horizontal axis.
2. Trace the curve as $x$ approaches $a$ from the left and from the right.
3. Observe what $y$-value the graph is approaching.
4. If both sides approach the same $y$-value, that’s the limit.

Key graphical cues:

• A hole (open circle) often indicates the function value is missing there, but the curve may still approach that point.
• A filled dot indicates the actual value $f(a)$.
• A jump means left and right approach different heights.
• A vertical asymptote suggests the function grows without bound as you approach $a$.

One-sided limits from a graph

If a graph is only defined on one side of $a$ (for instance, a square-root curve that starts at $a$), then you can only talk about the appropriate one-sided limit.

For example, if a graph begins at $x = 0$ and is only shown for $x \ge 0$, then $\lim_{x \to 0^-} f(x)$ might not make sense from the graph (and may not exist if the function isn’t defined for $x < 0$).

Estimating limits from a table

A table gives discrete samples, so you approximate a limit by choosing $x$-values closer and closer to $a$ from both sides.

A good process:

1. Use two sequences approaching $a$: one from below (like $1.9, 1.99, 1.999$) and one from above (like $2.1, 2.01, 2.001$).
2. Watch whether $f(x)$ appears to settle toward a single number.
3. Compare left-side and right-side trends.

What can go wrong with tables:

• If you use values that aren’t close enough, you may miss the true trend.
• Rounding can hide the approach.
• If the function changes rapidly near $a$, you might need much closer values.

Example 1: A limit from a graph with a hole

Suppose a graph shows a smooth curve approaching the point $\left(2, 5\right)$ but with an open circle at $\left(2, 5\right)$ and a filled dot at $\left(2, 1\right)$.

Then:

• The limit depends on what the curve approaches: $5$.
• The function value is the filled dot’s $y$-value: $f(2) = 1$.

So you would conclude:

$\lim_{x \to 2} f(x) = 5$

and

$f(2) = 1$

This is a classic “limit vs value” assessment.

Example 2: A limit from a table

You are given values near $x = 3$:

From the left, $f(x)$ seems to approach $7$; from the right, it also approaches $7$. So the best estimate is:

$\lim_{x \to 3} f(x) = 7$

Even if the table also listed $f(3) = 10$, the limit would still be $7$ because the surrounding behavior drives the limit.

Example 3: Detecting a non-existent limit from a table

If you see values approaching $2$ from the left give outputs near $4$, but values approaching $2$ from the right give outputs near $-1$, then you should conclude:

$\lim_{x \to 2} f(x)$ does not exist.

It’s not “undefined because of division by zero” or “undefined because there is a gap.” It fails specifically because the two one-sided limits disagree.

Exam Focus

• Typical question patterns:
  • Estimate $\lim_{x \to a} f(x)$ or one-sided limits from a graph with holes, jumps, or asymptotes.
  • Given a table, determine whether the limit exists and approximate it.
  • Compare $\lim_{x \to a} f(x)$ to $f(a)$ using graphical/table information.
• Common mistakes:
  • Reading $f(a)$ (the filled dot or the table entry at $a$) instead of the approached value.
  • Using only one side of $a$ and concluding a two-sided limit exists.
  • Declaring “DNE” too quickly when the function is undefined at $a$ even though the surrounding values clearly approach a number.

Determining Limits Using Algebraic Properties and Manipulation

Why algebra shows up in limits

Many limits on AP Calculus are designed so that direct substitution almost works, but you hit an indeterminate form such as $\frac{0}{0}$. Algebraic manipulation lets you rewrite the expression into an equivalent form (for $x$ near the target value) where the limit becomes clear.

This is not just trickery: it reflects the core limit idea that what matters is the behavior near the point. If two formulas are equal for all $x$ close to $a$ (even if one is not defined at $a$), they will have the same limit as $x \to a$.

The basic limit laws (algebraic properties)

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ (and the limits exist), then the following laws hold:

$\lim_{x \to a} (f(x) + g(x)) = L + M$

$\lim_{x \to a} (f(x) - g(x)) = L - M$

$\lim_{x \to a} (c f(x)) = cL$

$\lim_{x \to a} (f(x) g(x)) = LM$

$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$

This last one requires $M \ne 0$.

A hugely important special case is direct substitution: if $f(x)$ is a polynomial or a rational function with a nonzero denominator at $a$, then

$\lim_{x \to a} f(x) = f(a)$

because such functions are continuous at those points.

Indeterminate forms and what they mean

An indeterminate form is an algebraic output like $\frac{0}{0}$ that does not tell you the limit by itself. Different functions can produce $\frac{0}{0}$ but have completely different limits.

For example:

• $\frac{x}{x}$ gives $1$ for $x \ne 0$, so $\lim_{x \to 0} \frac{x}{x} = 1$.
• $\frac{x^2}{x} = x$ for $x \ne 0$, so $\lim_{x \to 0} \frac{x^2}{x} = 0$.

Both look like $\frac{0}{0}$ at $x = 0$, but the limits differ.

Core algebraic techniques

When direct substitution yields $\frac{0}{0}$, common manipulations include:

1) Factoring and canceling

If a common factor causes the zero in numerator and denominator, factor and cancel (but remember: canceling changes the expression at the point where the factor is zero, which is fine for limits).

Worked example

$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

Direct substitution gives $\frac{0}{0}$. Factor the numerator:

$x^2 - 9 = (x - 3)(x + 3)$

So for $x \ne 3$,

$\frac{x^2 - 9}{x - 3} = x + 3$

Now take the limit:

$\lim_{x \to 3} (x + 3) = 6$

So:

$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$

Misconception to avoid: canceling is not “illegal” here; it would be illegal only if you then claimed the original function is defined at $x = 3$. You’re using cancellation to understand behavior near $3$.

2) Rationalizing with conjugates

If radicals cause the indeterminate form, multiplying by a conjugate often simplifies.

Worked example

$\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$

Direct substitution gives $\frac{0}{0}$. Multiply by the conjugate:

$\frac{\sqrt{1 + x} - 1}{x} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}$

The numerator becomes a difference of squares:

$(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1) = (1 + x) - 1 = x$

So the expression simplifies (for $x \ne 0$) to:

$\frac{x}{x(\sqrt{1 + x} + 1)} = \frac{1}{\sqrt{1 + x} + 1}$

Now substitute $x = 0$:

$\lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{2}$

3) Simplifying complex fractions

Complex fractions often hide a factor you can cancel. A reliable approach is to combine into a single fraction or multiply numerator and denominator by the least common denominator.

Worked example

$\lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2}$

Combine the numerator:

$\frac{1}{x} - \frac{1}{2} = \frac{2 - x}{2x} = -\frac{x - 2}{2x}$

So the whole expression becomes:

$\frac{-\frac{x - 2}{2x}}{x - 2}$

Cancel $x - 2$ (for $x \ne 2$):

$-\frac{1}{2x}$

Now take the limit:

$\lim_{x \to 2} -\frac{1}{2x} = -\frac{1}{4}$

This type of limit is closely related to derivative definitions later, so it’s worth understanding the algebra.

Special trigonometric limit you should know

A foundational trig limit (used constantly later) is:

$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

On AP Calculus, you use it to evaluate many trig limits by algebraic rearrangement.

Example

$\lim_{x \to 0} \frac{\sin(5x)}{x}$

Rewrite to create the standard form:

$\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x}$

Now take the limit:

$\lim_{x \to 0} 5 \cdot \frac{\sin(5x)}{5x} = 5 \cdot 1 = 5$

A common mistake is to try to “cancel the sine” with the $x$. Instead, your goal is to match the known limit structure.

Exam Focus

• Typical question patterns:
  • Evaluate a limit that gives $\frac{0}{0}$ by factoring or rationalizing.
  • Use limit laws to break a complicated expression into simpler limits.
  • Evaluate trig limits by rewriting into a form involving $\lim_{x \to 0} \frac{\sin(x)}{x}$.
• Common mistakes:
  • Canceling terms that are not factors (for example, canceling across addition like canceling the $x$ in $x + 1$ with an $x$ in the denominator).
  • Forgetting to apply the conjugate correctly, leading to algebra errors.
  • Using the quotient law when the denominator’s limit is $0$ (you must address that issue first).

Squeeze Theorem

What the Squeeze Theorem says

The Squeeze Theorem is a way to find a limit when a function is trapped between two other functions whose limits you already know.

Formally, if for all $x$ near $a$ (possibly excluding $a$ itself) you have

$g(x) \le f(x) \le h(x)$

and

$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$

then

$\lim_{x \to a} f(x) = L$

Intuition: if $f(x)$ can’t go below $g(x)$ or above $h(x)$, and both bounding functions are forced toward the same value $L$, then $f(x)$ has no choice but to approach $L$ too.

Why it matters

The Squeeze Theorem is especially useful when:

• The function involves oscillation (like sine and cosine) multiplied by something shrinking to $0$.
• Algebraic simplification is messy or impossible.
• You can bound a complicated expression with simpler ones.

It also builds good calculus instincts: instead of trying to “compute exactly,” you reason about bounds and behavior.

How to apply it step by step

1. Identify a difficult function $f(x)$ whose limit you need.
2. Find two functions $g(x)$ and $h(x)$ such that $g(x) \le f(x) \le h(x)$ near the point.
3. Compute the limits of $g(x)$ and $h(x)$.
4. If both limits match, conclude the limit of $f(x)$ equals that same value.

The hardest step is usually step 2: choosing good “squeezing” functions.

Classic bounding fact with sine and cosine

A very common inequality is:

$-1 \le \sin(x) \le 1$

and similarly:

$-1 \le \cos(x) \le 1$

If you multiply all parts of an inequality by a positive expression, the inequality direction stays the same. If you multiply by a negative expression, the inequality reverses. On AP, you typically multiply by something nonnegative like $|x|$ or $x^2$ to avoid sign issues.

Example 1: Oscillation times something shrinking

Evaluate:

$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$

You cannot use direct substitution because $\sin\left(\frac{1}{x}\right)$ oscillates infinitely as $x \to 0$. But you can bound it:

$-1 \le \sin\left(\frac{1}{x}\right) \le 1$

Multiply by $x^2$ (which is always nonnegative):

$-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$

Now compute the limits of the bounds:

$\lim_{x \to 0} -x^2 = 0$

$\lim_{x \to 0} x^2 = 0$

Since both squeeze to $0$, the middle must also approach $0$:

$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

This is a perfect example of how limits can exist even when the function is wildly oscillatory.

Example 2: Using absolute value as a bound

Evaluate:

$\lim_{x \to 0} x \cos\left(\frac{1}{x}\right)$

We know $-1 \le \cos\left(\frac{1}{x}\right) \le 1$, but multiplying by $x$ is tricky because $x$ can be negative. A safer strategy is to use absolute value:

$\left|x \cos\left(\frac{1}{x}\right)\right| = |x| \cdot \left|\cos\left(\frac{1}{x}\right)\right|$

Since $\left|\cos\left(\frac{1}{x}\right)\right| \le 1$, you get

$0 \le \left|x \cos\left(\frac{1}{x}\right)\right| \le |x|$

And

$\lim_{x \to 0} 0 = 0$

$\lim_{x \to 0} |x| = 0$

So

$\lim_{x \to 0} \left|x \cos\left(\frac{1}{x}\right)\right| = 0$

which forces

$\lim_{x \to 0} x \cos\left(\frac{1}{x}\right) = 0$

The lesson: when signs might flip, bounding the absolute value is often the cleanest route.

What can go wrong with the Squeeze Theorem

• If your upper and lower bounds approach different limits, you cannot conclude anything.
• Your inequality must be true for $x$ sufficiently close to $a$ (usually on some punctured interval around $a$), not just for a few points.
• Be careful multiplying inequalities by expressions that can be negative; use $x^2$ or $|x|$ when possible.

Exam Focus

• Typical question patterns:
  • Evaluate limits involving oscillating trig functions like $\sin\left(\frac{1}{x}\right)$ or $\cos\left(\frac{1}{x}\right)$ multiplied by a power of $x$.
  • Justify a limit using an inequality setup (you may be asked to show the bounding step explicitly).
  • Decide whether a limit exists when oscillation is present.
• Common mistakes:
  • Forgetting that squeezing requires two bounds with the same limit.
  • Multiplying an inequality by $x$ without accounting for sign changes.
  • Trying to use direct substitution on oscillatory expressions and concluding “DNE” even when the shrinking factor forces the product to approach $0$.