Limits at Infinity

♾️ Limits at Infinity — End Behavior

Part 1 of 4 — What $\lim_{x \to \pm\infty}$ means

Topics in This Part

🔑 Why this matters: End behavior governs horizontal asymptotes — a foundational topic for both limits and integration later.

💡 What Does $\lim_{x \to \infty} f(x) = L$ Mean?

As $x$ gets arbitrarily large, $f(x)$ gets arbitrarily close to $L$.

We are not plugging $\infty$ into $f$ — $\infty$ isn't a number. We're asking: as $x$ marches off to the right, where is $f$ headed?

Same idea for $\lim_{x \to -\infty} f(x) = L$ — march to the left.

💡 The two ends are independent: $f$ might approach different limits as $x \to \infty$ and as $x \to -\infty$.

🔑 Horizontal Asymptotes

The line $y = L$ is a horizontal asymptote of $f$ if $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$.

A function can have:

• 0 horizontal asymptotes (e.g., $f(x) = x^2$ — both ends $\to +\infty$)
• 1 horizontal asymptote (e.g., $f(x) = e^x$ — left end $\to 0$, right end $\to \infty$, so $y = 0$ on the left only)
• 2 horizontal asymptotes (e.g., $f(x) = \arctan x$ — $-\pi/2$ on the left, $\pi/2$ on the right)

⚠️ A function can cross a horizontal asymptote — it just can't stay away from it forever.

📋 Quick Reference

End-Behavior Vocab 🎯

Compute Limits at Infinity 🧮

1) $\lim_{x \to \infty} \dfrac{5}{x} = ?$

2) $\lim_{x \to -\infty} e^x = ?$

3) $\lim_{x \to \infty} \arctan x = ?$ (use pi/2)

📊 Rational Functions at Infinity

Part 2 of 4 — Compare degrees

Topics in This Part

🔑 Why this matters: Almost every AP "limit at infinity" problem on a rational function is solved by the degree-comparison rule.

🔑 The Three-Case Rule

For $\lim_{x \to \pm\infty} \dfrac{p(x)}{q(x)}$ with degrees $\deg p = m$, $\deg q = n$:

💡 Memorize this. It's the single most useful rule in this section.

🛠️ Why It Works — Divide By Highest Power

Take $\lim_{x \to \infty} \dfrac{3x^2 + 5x}{2x^2 - 7}$. Divide top and bottom by $x^2$:

$\dfrac{3 + 5/x}{2 - 7/x^2}$

As $x \to \infty$, $5/x \to 0$ and $7/x^2 \to 0$:

$\dfrac{3 + 0}{2 - 0} = \dfrac{3}{2}$

The non-leading terms become negligible, leaving just the ratio of leading coefficients.

📝 Examples

Apply the Three-Case Rule 🎯

Compute 🧮

1) $\lim_{x \to \infty} \dfrac{8x^3 - 5}{4x^3 + x} = ?$

2) $\lim_{x \to \infty} \dfrac{3x + 1}{x^2 - 7} = ?$

3) $\lim_{x \to -\infty} \dfrac{6x^2 - 11}{2x^2 + 5} = ?$

√ Radicals at Infinity

Part 3 of 4 — Be careful with $|x|$ when $x \to -\infty$

Topics in This Part

🔑 Why this matters: A square root contributes "half a degree." And the sign on $x$ matters when $x$ is negative.

📐 Effective Degree

For end behavior, $\sqrt{x^2}$ behaves like $x$ in size — it has "effective degree 1." More generally, $\sqrt{x^{2k}}$ has effective degree $k$.

Example. $\lim_{x \to \infty} \dfrac{\sqrt{x^2 + 1}}{x + 1}$.

• Numerator effective degree 1; denominator degree 1 → equal.
• Leading behavior: numerator $\sim x$, denominator $\sim x$. Ratio $\to 1$.

So this limit is $1$ as $x \to \infty$.

⚠️ The $\sqrt{x^2} = |x|$ Trap

The identity is: $\sqrt{x^2} = |x| = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}$

So when $x \to -\infty$, $\sqrt{x^2} = -x$ (positive), not $x$. This sign change matters.

Worked example. $\lim_{x \to -\infty} \dfrac{\sqrt{x^2 + 1}}{x + 1}$.

• Divide top and bottom by $|x| = -x$ (since $x \to -\infty$):

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

• As $x \to -\infty$: numerator $\to \sqrt{1} = 1$, denominator $\to -1$.

Answer: $\dfrac{1}{-1} = -1$.

💡 Compare: as $x \to +\infty$ the same expression gives $+1$. The two ends differ in sign.

🚀 Shortcut: Look at Leading Behavior

For $\sqrt{ax^2 + \ldots}$ as $x \to \pm\infty$:

• $x \to +\infty$: $\sqrt{ax^2 + \ldots} \sim \sqrt{a}\,x$
• $x \to -\infty$: $\sqrt{ax^2 + \ldots} \sim -\sqrt{a}\,x$ (because we need a positive output and $x$ is negative)

This shortcut handles most AP problems quickly.

Sign-Aware Limits 🎯

Practice 🧮

1) $\lim_{x \to \infty} \dfrac{\sqrt{9x^2 + x}}{x + 5} = ?$

2) $\lim_{x \to -\infty} \dfrac{\sqrt{9x^2 + x}}{x + 5} = ?$

🌐 Other Cases — Exponentials, Logs, and Oscillation

Part 4 of 4 — Beyond rationals

Topics in This Part

🔑 Why this matters: Many limit-at-infinity problems involve $e^x$, $\ln x$, or oscillating functions. Each has its own rules.

🚀 Exponentials Beat Polynomials

For any positive $n$: $e^x$ grows faster than $x^n$. So $\lim_{x \to \infty} \dfrac{x^n}{e^x} = 0, \qquad \lim_{x \to \infty} \dfrac{e^x}{x^n} = +\infty.$

Example. $\lim_{x \to \infty} \dfrac{x^{100}}{e^x} = 0$. (Even with a huge polynomial, the exponential wins.)

For $x \to -\infty$: $e^x \to 0$, so things like $x^2 e^x \to 0$ (still exponential dominates the polynomial).

🐢 Logs Lose to Polynomials

For any positive $n$: $\ln x$ grows slower than $x^n$.

$\lim_{x \to \infty} \dfrac{\ln x}{x^n} = 0, \qquad \lim_{x \to \infty} \dfrac{x^n}{\ln x} = +\infty.$

Growth hierarchy (slowest to fastest, going to $+\infty$): $\ln x \;\ll\; x^{1/2} \;\ll\; x \;\ll\; x^2 \;\ll\; e^x$

🚫 When the Limit at Infinity DNE

Some functions don't settle to a value or to $\pm\infty$ at infinity:

• $\sin x$, $\cos x$ as $x \to \infty$ — oscillate forever between $-1$ and $1$. DNE.
• $x \sin x$ — amplitude grows, oscillates between $\pm\infty$. DNE.

But:

• $\dfrac{\sin x}{x} \to 0$ as $x \to \infty$ (Squeeze Theorem: bounded numerator, growing denominator).

💡 Bounded over growing gives 0. Growing × bounded oscillation fails to converge.

Beyond Rationals 🎯

Exit Check ✅