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Limits · Mini-Quiz Q5

Limit at infinity, equal degrees — the ratio of leading coefficients is the whole answer

Question 5 (1 pt)

Evaluate

$$\lim_{x \to \infty} \frac{x^3 - 2x^2 + x - 3}{3x^3 - 2x + 1}$$

Answer: $\dfrac{1}{3}$

1 · The intuition

Intuition

Look at the degrees first, the way you did in Q4:

They're the same. That changes everything.

When top and bottom grow at the same rate, the fraction settles down to a single finite number — the ratio of how fast they grow. Like two trains at the same speed: the gap between them stays constant. For polynomials, the "growth rate" of a degree-3 polynomial is set by its leading coefficient, so the limit is just leading coefficient of top ÷ leading coefficient of bottom.

Here: $\frac{1}{3}$.

2 · The rule (with derivation)

Formula — equal degrees

For $\displaystyle\lim_{x \to \infty} \frac{a_n x^n + \text{lower}}{b_n x^n + \text{lower}}$ with $a_n, b_n \neq 0$ and the same degree $n$ on top and bottom:

$$\lim_{x \to \infty} \frac{a_n x^n + \cdots}{b_n x^n + \cdots} = \frac{a_n}{b_n}$$

Why? Factor out the highest power on both sides:

$$\frac{x^n\!\left(a_n + \frac{a_{n-1}}{x} + \cdots\right)}{x^n\!\left(b_n + \frac{b_{n-1}}{x} + \cdots\right)} = \frac{a_n + \frac{a_{n-1}}{x} + \cdots}{b_n + \frac{b_{n-1}}{x} + \cdots}$$

As $x \to \infty$, every $\frac{1}{x}$ term vanishes, leaving $\frac{a_n}{b_n}$.

3 · The worked solution

Step-by-step

Step 1 — Identify the leading terms.

Step 2 — Factor $x^3$ out of both.

$$\frac{x^3 - 2x^2 + x - 3}{3x^3 - 2x + 1} = \frac{x^3\!\left(1 - \frac{2}{x} + \frac{1}{x^2} - \frac{3}{x^3}\right)}{x^3\!\left(3 - \frac{2}{x^2} + \frac{1}{x^3}\right)}$$

Step 3 — Cancel $x^3$.

$$= \frac{1 - \frac{2}{x} + \frac{1}{x^2} - \frac{3}{x^3}}{3 - \frac{2}{x^2} + \frac{1}{x^3}}$$

Step 4 — Take the limit. All the $\frac{1}{x^k}$ terms go to 0:

$$= \frac{1 - 0 + 0 - 0}{3 - 0 + 0} = \frac{1}{3}$$

Answer: $\dfrac{1}{3}$.

4 · Why the other choices are wrong

Distractor analysis
ChoiceVerdictWhy
$\infty$ Wrong That's the Q4 trap — top degree strictly greater than bottom degree. Here they're equal, so the limit is a finite number, not infinity.
3 Wrong You got the ratio upside down. The leading coefficient on top is $1$, the one on the bottom is $3$, so the ratio is $1/3$, not $3/1$. (The fraction 3 alone is what you'd get if you had $\frac{3x^3}{x^3}$.)
0 Wrong That's the case where the bottom grows faster (degree strictly bigger). Here they're equal, so the limit is nonzero and finite.
$\dfrac{1}{3}$ Correct Equal degrees → ratio of leading coefficients = $\frac{1}{3}$.

5 · The big picture — the three-outcome flowchart

The complete rule for $\lim_{x \to \infty} \frac{p(x)}{q(x)}$
Top degree vs bottom degree Limit What it means
Top < bottom $0$ Bottom grows faster → fraction shrinks to 0
Top = bottom $\dfrac{a_n}{b_n}$ (ratio of leading coefficients) Both grow at same rate → settle to a constant
Top > bottom $\pm\infty$ Top grows faster → blows up

Q4 = middle row of the previous lesson (top wins → $\infty$). Q5 = middle row of this lesson (equal → ratio of leading coefficients). Once you can recognize which case you're in, the answer is mechanical.

6 · Common traps

Watch out

7 · Check your understanding

Practice 1: $\displaystyle\lim_{x \to \infty} \frac{4x^2 - 1}{2x^2 + 5x}$

Both degree 2. Equal → ratio of leading coefficients.

$\frac{4}{2} = 2$.

Answer: $2$.

Practice 2: $\displaystyle\lim_{x \to \infty} \frac{6x^4 + x^2}{2x^4 - 3x^3}$

Both degree 4. Equal → ratio of leading coefficients.

$\frac{6}{2} = 3$.

Answer: $3$.

Practice 3: $\displaystyle\lim_{x \to \infty} \frac{9x^2 - 4x}{3x^3 + 1}$

Top degree 2, bottom degree 3. Bottom wins → $0$.

(Don't fall for the trap of "the leading coefficient ratio is $9/3 = 3$" — that's only valid when degrees are equal.)

Answer: $0$.

8 · Takeaways

← Mini-Quiz Q4 All Graphical Limits questions Main pretest →

built by Aida · 2026-06-25