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Limits · Mini-Quiz Q4

Limit at infinity, polynomial-over-polynomial, top degree wins

Question 4 (1 pt)

Evaluate

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

Answer: $\infty$

1 · The intuition

Intuition

"$x \to \infty$" means "$x$ grows without bound." The question becomes: which part of this fraction blows up faster, the top or the bottom?

Look at the degrees:

When you divide something that grows like $x^2$ by something that grows like $x$, you get something that grows like $x$. So the limit runs off to $\infty$.

2 · The rule

Formula — limits at infinity for rational functions

For a rational function $\dfrac{p(x)}{q(x)}$ where $p, q$ are polynomials, only the highest-degree terms matter as $x \to \infty$. The result depends on the degree comparison:

Case Limit Why
$\deg p < \deg q$ $0$ Bottom grows faster → fraction shrinks to 0
$\deg p = \deg q$ $\dfrac{a_n}{b_n}$ Ratio of leading coefficients
$\deg p > \deg q$ $\pm\infty$ Top grows faster → fraction blows up

Q4 is the third row: degree 2 > degree 1, so the limit is $\pm\infty$. Since the leading coefficient $3$ is positive, it's $+\infty$.

3 · The worked solution

Step-by-step

Step 1 — Compare degrees. Top is degree 2, bottom is degree 1. Top wins, so the limit is $\pm\infty$.

Step 2 — Verify by dividing every term in the top by the bottom.

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

Step 3 — Take the limit term-by-term.

$$\lim_{x \to \infty} \left(3x - 2 + \frac{5}{x}\right) = \underbrace{3x}_{\to\, \infty} \;-\; \underbrace{2}_{\to\, 2} \;+\; \underbrace{\frac{5}{x}}_{\to\, 0}$$

The $-2$ and the $\frac{5}{x}$ become irrelevant. The $3x$ term goes to $\infty$.

Answer: $\infty$.

4 · Why the other choices are wrong

Distractor analysis
ChoiceVerdictWhy
3 Wrong That's the answer when degrees are equal and you take the ratio of leading coefficients (Q5). Here, the top is degree 2, the bottom is degree 1 — the answer can't be a single number.
0 Wrong That's the answer when the bottom degree is bigger than the top (Q2-style). Here, the top wins.
DNE Wrong The limit doesn't "not exist" in the pathological sense — it just grows without bound. In calculus, we do say the limit is $\infty$; DNE is reserved for limits that approach different finite values from each side (jump discontinuities, Q5/Q9 of the main pretest).
$\infty$ Correct Top degree (2) > bottom degree (1) → top wins → $\infty$.

5 · Common traps

Watch out

6 · Check your understanding

Practice 1: $\displaystyle\lim_{x \to \infty} \frac{5x^3 + x}{x^2 + 1}$

Top degree 3, bottom degree 2. Top wins. The expression behaves like $\frac{5x^3}{x^2} = 5x \to \infty$.

Answer: $\infty$.

Practice 2: $\displaystyle\lim_{x \to \infty} \frac{7x^2 - 4}{2x^2 + x}$

Top and bottom both degree 2. Equal degrees → ratio of leading coefficients.

$\frac{7x^2}{2x^2} = \frac{7}{2}$.

Answer: $\frac{7}{2}$.

Practice 3: $\displaystyle\lim_{x \to \infty} \frac{4x - 1}{x^3 + 7}$

Top degree 1, bottom degree 3. Bottom wins → fraction shrinks to 0.

Answer: $0$.

7 · Takeaways

← Mini-Quiz Q3 All Graphical Limits questions Mini-Quiz Q5 →

built by Aida · 2026-06-25