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Limits · Pretest Q2

Limits at infinity of rational functions — the degree-comparison rule

Question 2 (1 pt)

Evaluate $\displaystyle\lim_{x \to \infty} \frac{4x^2 - 2x + 1}{x - 1}$.

0
4
DNE
$\infty$

1 · The intuition

Intuition

This is a limit at infinity — we want to know what happens to the function as $x$ gets very, very large. The trick is to ignore the noise (the smaller terms) and just compare the highest-degree terms of the top and bottom.

It's like asking: "If I walk further and further away, does a flashlight beam in a foggy field converge, diverge, or stay steady?" The answer depends on how the beam spreads vs. how the fog thins — in math terms, the ratio of leading terms.

2 · The rule (rational functions at infinity)

Formula — degree comparison

For a rational function $\dfrac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials:

Compare degrees	Limit as $x \to \pm\infty$
$\deg p < \deg q$	Limit $= 0$ (denominator wins, function dies off)
$\deg p = \deg q$	Limit $= $ ratio of leading coefficients (a finite number)
$\deg p > \deg q$	Limit $= \pm\infty$ (numerator wins, function blows up)

For our problem: $\deg(\text{numerator}) = 2$ and $\deg(\text{denominator}) = 1$, so we're in the third row. The limit is $\pm\infty$. The sign depends on the leading coefficients' signs and whether $x \to +\infty$ or $x \to -\infty$.

3 · Worked solution

Worked solution

For $\displaystyle\lim_{x \to \infty} \frac{4x^2 - 2x + 1}{x - 1}$:

Step 1 — Identify the degrees.

Numerator: $4x^2 - 2x + 1$ → degree 2, leading coefficient $+4$
Denominator: $x - 1$ → degree 1, leading coefficient $+1$

Degree of numerator (2) > degree of denominator (1) → limit is $\pm\infty$.

Step 2 — Determine the sign.

The dominant behavior as $x \to +\infty$ is:

$$\frac{4x^2}{x} = 4x$$

Both leading coefficients are positive and $x \to +\infty$, so $4x \to +\infty$. The smaller terms ($-2x$, $+1$, $-1$) become negligible. So the function grows to $+\infty$.

Step 3 — (Optional) Divide by the highest power in the denominator. This is the formal technique that makes the limit rigorous:

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

As $x \to \infty$, the $\dfrac{1}{x}$ terms vanish:

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

So the answer is:

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

Numerical sanity check. Plug in large $x$ values:

$x$	$(4x^2 - 2x + 1) / (x - 1)$
$10$	$42.3$
$100$	$402.0$
$1{,}000$	$4{,}002.0$
$10{,}000$	$40{,}002.0$
$1{,}000{,}000$	$4{,}000{,}002.0$

Notice the pattern: value is approximately $4x$ for large $x$. Growing without bound — confirms $+\infty$.

4 · Why this matters

Deeper "why"

The degree-comparison rule is one of the workhorses of calculus and statistics. You'll see it in:

End behavior of polynomials — sketching graphs in calc I
Asymptotic analysis — the basis of "big-O" notation in CS / algorithm analysis
Tail behavior of distributions — how fast a probability density dies off as $x \to \infty$ (Cauchy vs. Normal, etc.)
Convergence of series — $\sum 1/n^p$ converges when $p > 1$, which is the same rule in disguise

The intuition: the highest-degree term dominates the long run. Everything else is just noise in the limit.

5 · Common traps

Watch out

"DNE" is the wrong trap. "DNE" is for limits that genuinely don't exist (e.g., $\lim_{x \to 0} \sin(1/x)$). A limit going to infinity is a real limit — it just happens to be $\pm\infty$. So if you can see the function growing without bound, the answer is $\infty$, not DNE.
"$0$" is the wrong trap. $0$ would be the answer if the denominator grew faster than the numerator (e.g., $\frac{x}{x^2}$ as $x \to \infty$ gives $0$).
"$4$" is the wrong trap. That's the leading coefficient ratio — that would be the answer if both degrees were equal (e.g., $\frac{4x^2 + \ldots}{x^2 + \ldots} \to 4$ as $x \to \infty$).
Watch the sign when $x \to -\infty$. The same expression as $x \to -\infty$ gives $-\infty$ (because $4x$ flips sign for negative $x$). Always check whether you're going to $+\infty$ or $-\infty$.

6 · Check your understanding

Practice

Try these — use the degree-comparison rule to predict the answer, then mentally verify:

$\displaystyle\lim_{x \to \infty} \frac{3x + 5}{x^2 - 7}$ → degrees 1 vs 2, numerator loses
$\displaystyle\lim_{x \to \infty} \frac{2x^2 + x}{5x^2 - 3}$ → degrees 2 vs 2, ratio of leading coefficients
$\displaystyle\lim_{x \to -\infty} \frac{6x^3 - 1}{2x^2 + 9x}$ → degrees 3 vs 2, numerator wins — but watch the sign!

Show answers

$0$ — denominator grows faster, function dies off.
$2/5$ — equal degrees, ratio of leading coefficients is $2/5$.
$-\infty$ — numerator wins (degree 3 > 2), and the dominant term is $6x^3 / 2x^2 = 3x \to -\infty$ as $x \to -\infty$.

7 · What's next

The degree-comparison rule is a special case. For the full limits toolkit, the main tools are:

Direct substitution (the easy one)
Factoring (for $0/0$ forms)
Multiplying by the conjugate (for $\infty - \infty$ forms with radicals)
Trig identities and the squeeze theorem
L'Hôpital's rule (for tricky $0/0$ or $\infty/\infty$ forms)
Limits at infinity (this lesson)

Send the next pretest problem, or tell me which technique to drill next.

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