Limits Involving Infinity (AP Calculus BC Unit 1 Study Notes)

Infinite Limits and Vertical Asymptotes

What an infinite limit means

An infinite limit happens when the function values grow without bound (upward or downward) as the input gets close to some number. Instead of approaching a finite height, the graph shoots up or down near a particular vertical line.

Formally, you’ll see statements like:

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

This does not mean the limit “equals infinity” as a real number. It means: by choosing $x$ sufficiently close to $a$ (but not equal to $a$), you can make $f(x)$ as large and positive as you want.

Similarly:

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

means $f(x)$ becomes very large in magnitude and negative.

A key idea: infinite limits are still about behavior near $a$, not at $a$. The function might be undefined at $a$ (common), but it could even be defined there and still have an infinite limit if the nearby behavior blows up.

Why this matters

Infinite limits are the bridge between limits and asymptotes. They help you:

• Describe graph behavior near “blow-up” points.
• Identify and interpret vertical asymptotes, which are important for sketching graphs and analyzing rational functions.
• Understand discontinuities that are not removable (you can’t “patch” the hole with one point).

In later calculus topics, vertical asymptotes also show up when you analyze improper integrals, so building a solid intuition here pays off.

One-sided infinite limits (the most common situation)

Very often, the function behaves differently depending on whether you approach $a$ from the left or from the right. That’s why one-sided limits matter.

You might see:

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

and

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

This means: from the left the graph shoots up, and from the right it shoots down.

If the left-hand and right-hand behaviors don’t match, then the two-sided limit

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

does not exist (even though both sides “blow up” in some sense).

How infinite limits create vertical asymptotes

A vertical asymptote is a vertical line that the graph approaches as $x$ approaches some value. In AP Calculus language, the line

$x = a$

is a vertical asymptote of $f(x)$ if at least one of these is true:

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

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

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

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

So you do not need both sides to blow up in the same direction. One-sided blow-up is enough to declare a vertical asymptote.

A helpful mental picture: near a vertical asymptote, the graph behaves like it is being “pulled” toward the vertical line, but its $y$-values are flying off to extremely large positive or negative values.

How to determine whether the limit is $\infty$ or $-\infty$

When you have expressions like rational functions (fractions), the sign is everything.

A common structure is:

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

As $x \to a$, the denominator approaches $0$. But from the left and right it approaches $0$ with different signs:

• If $x \to a^-$ then $x-a$ is a tiny negative number, so $\frac{1}{x-a}$ is a large negative number.
• If $x \to a^+$ then $x-a$ is a tiny positive number, so $\frac{1}{x-a}$ is a large positive number.

Squaring changes this because squares are nonnegative:

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

Now, from either side, $(x-a)^2$ is a tiny positive number, so the fraction becomes large and positive on both sides.

A quick sign-analysis routine (works extremely well)

To evaluate something like

$\lim_{x \to a^-} \frac{g(x)}{h(x)}$

when $h(x) \to 0$:

1. Figure out the sign of the numerator $g(x)$ near $a$.
2. Figure out the sign of the denominator $h(x)$ near $a$ (left side or right side matters!).
3. Combine signs to decide whether the fraction goes to $\infty$ or $-\infty$.

Example 1: One-sided infinite limits

Evaluate the one-sided limits:

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

$\lim_{x \to 0^+} \frac{x+1}{x}$

Step 1: Determine signs near $0$.

• Near $x = 0$, the numerator $x+1$ is close to $1$, so it’s positive on both sides.
• The denominator $x$ is negative for $x < 0$ and positive for $x > 0$.

Left-hand limit: positive divided by negative gives negative, with denominator approaching $0$ in magnitude. So:

$\lim_{x \to 0^-} \frac{x+1}{x} = -\infty$

Right-hand limit: positive divided by positive gives positive, with denominator approaching $0$ in magnitude. So:

$\lim_{x \to 0^+} \frac{x+1}{x} = \infty$

Because the one-sided limits are not the same (one is $-\infty$ and the other is $\infty$), the two-sided limit does not exist.

Also, $x = 0$ is a vertical asymptote.

Example 2: Infinite limit with same behavior on both sides

Evaluate:

$\lim_{x \to 2} \frac{5}{(x-2)^2}$

As $x \to 2$, $(x-2)^2 \to 0$ but stays positive (except at $2$ itself). So the fraction grows without bound positively:

$\lim_{x \to 2} \frac{5}{(x-2)^2} = \infty$

This tells you the graph shoots upward on both sides of $x = 2$, and $x = 2$ is a vertical asymptote.

Example 3: Detecting vertical asymptotes vs removable discontinuities

Consider:

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

Factoring the numerator:

$x^2 - 1 = (x-1)(x+1)$

So for $x \ne 1$:

$f(x) = x+1$

As $x \to 1$, the simplified form approaches $2$, a finite number. So:

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

This is not an infinite limit, so there is no vertical asymptote at $x = 1$. Instead, the original function has a hole (a removable discontinuity) because the factor causing the denominator to be zero cancels.

This comparison is crucial: a denominator of zero does not automatically guarantee a vertical asymptote; you must check whether the factor cancels.

What can go wrong (common conceptual pitfalls)

Students often confuse these three ideas:

• Undefined at $a$ (function value missing)
• Limit exists at $a$ (behavior approaches a single number)
• Vertical asymptote at $a$ (behavior becomes infinite)

A function can be undefined at $a$ and still have a perfectly finite limit (a hole). Or it can be defined at $a$ but have an infinite limit because of nearby behavior. Always focus on what happens as $x$ approaches.

Exam Focus

• Typical question patterns
  • Determine whether $x = a$ is a vertical asymptote by evaluating one-sided limits such as $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$.
  • Evaluate infinite limits of rational functions by sign analysis near where the denominator is $0$.
  • Distinguish between a vertical asymptote and a removable discontinuity by factoring and simplifying.
• Common mistakes
  • Declaring a vertical asymptote just because the denominator is $0$ at $x=a$ (you must check for cancellation).
  • Assuming $\lim_{x \to a} f(x) = \infty$ requires both sides to go to $\infty$ (one-sided infinite behavior is enough for a vertical asymptote).
  • Losing the sign when approaching from the left vs the right (especially with odd powers like $x-a$).

Limits at Infinity and Horizontal Asymptotes

What “limit at infinity” means

A limit at infinity describes what happens to a function’s values as $x$ becomes very large (either positively or negatively). Instead of zooming in near a point, you’re looking at the function’s “end behavior.”

You’ll see:

$\lim_{x \to \infty} f(x)$

and

$\lim_{x \to -\infty} f(x)$

If this limit equals a finite number $L$, it means that as you go far to the right (or left) on the graph, the function values get closer and closer to $L$.

Why this matters

Limits at infinity help you:

• Predict long-run behavior without graphing every detail.
• Identify horizontal asymptotes, which guide accurate graph sketches.
• Model real situations where outputs level off (for example, a concentration approaching a saturation level, or a diminishing effect where increases in input barely change the output).

Horizontal asymptotes: the limit connection

A horizontal asymptote is a horizontal line that the graph approaches as $x$ goes to $\infty$ or $-\infty$.

If

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

then the line

$y = L$

is a horizontal asymptote (to the right).

If

$\lim_{x \to -\infty} f(x) = M$

then

$y = M$

is a horizontal asymptote (to the left).

Important: it’s possible to have two different horizontal asymptotes—one for $x \to \infty$ and a different one for $x \to -\infty$.

Also important: a horizontal asymptote is not a “barrier.” A graph can cross a horizontal asymptote.

The main skill: evaluating limits at infinity for rational functions

In Unit 1, the most common functions are rational functions:

$f(x) = \frac{p(x)}{q(x)}$

where $p(x)$ and $q(x)$ are polynomials.

For rational functions, end behavior is dominated by the highest power (the leading term). The key comparison is the degree of the numerator vs the degree of the denominator.

Let the leading terms be:

$p(x) \approx a x^n$

$q(x) \approx b x^m$

Then for large $|x|$, the fraction behaves like:

$\frac{p(x)}{q(x)} \approx \frac{a x^n}{b x^m} = \frac{a}{b} x^{n-m}$

From here you get three major cases.

Case 1: Degree numerator < degree denominator

If $n < m$, then $x^{n-m}$ is like $\frac{1}{x^{m-n}}$, which goes to $0$ as $x \to \pm\infty$.

So:

$\lim_{x \to \infty} \frac{p(x)}{q(x)} = 0$

and typically also

$\lim_{x \to -\infty} \frac{p(x)}{q(x)} = 0$

This gives horizontal asymptote $y = 0$.

Case 2: Degree numerator = degree denominator

If $n = m$, the powers of $x$ essentially cancel in the end behavior, and the limit is the ratio of leading coefficients:

$\lim_{x \to \infty} \frac{p(x)}{q(x)} = \frac{a}{b}$

and similarly for $x \to -\infty$.

So the horizontal asymptote is:

$y = \frac{a}{b}$

Case 3: Degree numerator > degree denominator

If $n > m$, then $x^{n-m}$ grows without bound in magnitude as $x \to \infty$ (and may go to $\infty$ or $-\infty$ depending on parity and signs). In this case, there is no horizontal asymptote.

(There can be other asymptotes like slant/oblique when the degree is exactly one higher, but the essential Unit 1 takeaway is: the limit at infinity is not finite, so you do not get a horizontal asymptote.)

A compact comparison table (rational functions)

Example 1: Degree numerator less than degree denominator

Evaluate:

$\lim_{x \to \infty} \frac{4x-7}{2x^2+1}$

Here, the numerator has degree $1$ and the denominator has degree $2$, so the fraction should approach $0$.

If you want to see it algebraically, divide top and bottom by $x^2$:

$\frac{4x-7}{2x^2+1} = \frac{\frac{4}{x}-\frac{7}{x^2}}{2+\frac{1}{x^2}}$

As $x \to \infty$, the terms $\frac{4}{x}$, $\frac{7}{x^2}$, and $\frac{1}{x^2}$ all go to $0$, so:

$\lim_{x \to \infty} \frac{4x-7}{2x^2+1} = \frac{0-0}{2+0} = 0$

Horizontal asymptote: $y = 0$.

Example 2: Same degree (leading coefficient ratio)

Evaluate:

$\lim_{x \to -\infty} \frac{3x^2-5}{2x^2+7x}$

Both numerator and denominator have degree $2$. The leading coefficients are $3$ and $2$, so the limit is:

$\lim_{x \to -\infty} \frac{3x^2-5}{2x^2+7x} = \frac{3}{2}$

If you divide by $x^2$ to justify it:

$\frac{3x^2-5}{2x^2+7x} = \frac{3-\frac{5}{x^2}}{2+\frac{7}{x}}$

As $x \to -\infty$, $\frac{5}{x^2} \to 0$ and $\frac{7}{x} \to 0$, so the expression approaches $\frac{3}{2}$.

Horizontal asymptote: $y = \frac{3}{2}$.

Example 3: A non-rational example using an algebra trick (conjugates)

Not every AP limit at infinity is a polynomial ratio. A classic skill is rewriting expressions to avoid an “infinity minus infinity” form.

Evaluate:

$\lim_{x \to \infty} \left(\sqrt{x^2+1} - x\right)$

Direct substitution isn’t meaningful: both $\sqrt{x^2+1}$ and $x$ grow large, so it looks like $\infty - \infty$ (an indeterminate form).

Step 1: Multiply by the conjugate.

$\sqrt{x^2+1} - x$

Multiply numerator and denominator by the conjugate $\sqrt{x^2+1} + x$ (you’re really multiplying by $1$):

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

Step 2: Use difference of squares.

$(\sqrt{x^2+1})^2 - x^2 = (x^2+1) - x^2 = 1$

So the expression simplifies to:

$\frac{1}{\sqrt{x^2+1} + x}$

Step 3: Evaluate the limit.
As $x \to \infty$, the denominator grows without bound, so the whole fraction goes to $0$:

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

This example matters because it trains you to recognize and fix indeterminate forms without using later-unit tools like L’Hospital’s Rule.

Different behavior as $x \to \infty$ vs $x \to -\infty$

Some functions approach different values on different ends. A simple example is:

$f(x) = \frac{|x|}{x}$

For large positive $x$, $|x| = x$ so $f(x) = 1$. For large negative $x$, $|x| = -x$ so $f(x) = -1$.

So:

$\lim_{x \to \infty} \frac{|x|}{x} = 1$

$\lim_{x \to -\infty} \frac{|x|}{x} = -1$

This produces two horizontal asymptotes: $y = 1$ on the right and $y = -1$ on the left.

What can go wrong (common conceptual pitfalls)

• Thinking “horizontal asymptote” means the function never crosses it. A function can cross a horizontal asymptote; the asymptote describes end behavior, not a restriction.
• Forgetting to check both ends. Even if $\lim_{x \to \infty} f(x)$ exists, $\lim_{x \to -\infty} f(x)$ might be different.
• For rational functions, using non-leading terms to decide the limit. As $|x|$ gets large, lower-degree terms become insignificant compared to the highest power.

Exam Focus

• Typical question patterns
  • Compute $\lim_{x \to \infty} \frac{p(x)}{q(x)}$ or $\lim_{x \to -\infty} \frac{p(x)}{q(x)}$ by comparing degrees or dividing by the highest power of $x$.
  • Identify the horizontal asymptote(s) from limits at infinity, sometimes asking separately for $x \to \infty$ and $x \to -\infty$.
  • Evaluate a limit at infinity that gives an indeterminate form like $\infty - \infty$ by algebraic rewriting (often using conjugates).
• Common mistakes
  • Saying a horizontal asymptote exists whenever a rational function is present (if degree numerator is larger, the limit is not finite).
  • Dividing by the wrong power of $x$ (you should divide by the highest power present in the denominator or the overall highest power you want to factor out).
  • Treating $\lim_{x \to \infty} f(x) = L$ as if it means $f(x) = L$ for large $x$; the function only gets close to $L$, it doesn’t become exactly $L$.