← All courses › Calculus Refresher › Limits › Introduction (Pretest) › Q3

Limits · Pretest Q3

The flip side of Q2: when the denominator wins, the limit is zero

Question 3 (1 pt)

Evaluate $\displaystyle\lim_{x \to \infty} \frac{x-1}{x^2+2x+5}$.

1 · The intuition (recap from Q2)

Intuition

This is the mirror image of Q2. Same setup — a rational function, limit at infinity — but the degree relationship is flipped:

Q2: numerator degree 2, denominator degree 1 → top wins → $\infty$
Q3 (here): numerator degree 1, denominator degree 2 → bottom wins → $0$

The intuition: the denominator's $x^2$ eventually dwarfs the numerator's $x$. Big number on the bottom of a fraction = small result. The function is being drowned by its own denominator.

2 · The rule (same as Q2, different row)

Formula — degree comparison

For a rational function $\dfrac{p(x)}{q(x)}$ as $x \to \infty$:

Compare degrees	Limit as $x \to \infty$	Example
$\deg p < \deg q$	Limit $= 0$ (denominator wins)	$\frac{x}{x^2} \to 0$
$\deg p = \deg q$	Limit $= $ ratio of leading coefficients	$\frac{2x^2}{5x^2} \to 2/5$
$\deg p > \deg q$	Limit $= \pm\infty$ (numerator wins)	$\frac{x^2}{x} \to \infty$

For Q3: $\deg(\text{num}) = 1$ and $\deg(\text{den}) = 2$, so we're in row 1. Limit $= 0$.

3 · Worked solution

Worked solution

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

Step 1 — Identify the degrees.

Numerator: $x - 1$ → degree 1, leading coefficient $+1$
Denominator: $x^2 + 2x + 5$ → degree 2, leading coefficient $+1$

Degree of numerator (1) < degree of denominator (2) → limit is $0$.

Step 2 — (Optional) Divide by the highest power in the denominator. The formal technique:

$$\frac{x-1}{x^2+2x+5} = \frac{\dfrac{x-1}{x^2}}{\dfrac{x^2+2x+5}{x^2}} = \frac{\dfrac{1}{x} - \dfrac{1}{x^2}}{1 + \dfrac{2}{x} + \dfrac{5}{x^2}}$$

As $x \to \infty$, every $\dfrac{1}{x^k}$ term vanishes:

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

So the answer is:

$\displaystyle\lim_{x \to \infty} \frac{x-1}{x^2+2x+5} = 0$

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

$x$	$(x-1) / (x^2+2x+5)$
$10$	$0.0720$
$100$	$0.0097$
$1{,}000$	$0.00100$
$10{,}000$	$0.000100$
$1{,}000{,}000$	$0.000001$

Notice the pattern: value is approximately $1/x$ for large $x$. Heading toward $0$ — confirms the answer.

4 · Q2 vs. Q3 — the symmetric pair

The connection

Q2 and Q3 are inverses of each other. Put them side by side:

	Q2	Q3
Expression	$\frac{4x^2 - 2x + 1}{x - 1}$	$\frac{x - 1}{x^2 + 2x + 5}$
Num degree	2	1
Den degree	1	2
Limit	$\infty$	$0$

Notice: if you invert Q3 (flip top and bottom), you get a function with the same shape as Q2 — and the limit flips from $0$ to $\infty$. The reciprocal property:

$$\lim_{x \to \infty} \frac{1}{f(x)} = \frac{1}{\lim f(x)} \quad \text{(when the limit is nonzero finite)}$$

This is a useful shortcut: if you can solve one, you can solve the other by flipping.

5 · Common traps

Watch out

"The answer is $\infty$." This is the most common wrong answer — confusing Q3 with Q2. Always check: which degree is bigger, the top or the bottom? Top wins → $\infty$. Bottom wins → $0$.
"The answer is $1$." Easy mistake: $\frac{x}{x^2}$ "feels like" it should approach $1$, but it actually approaches $0$. The denominator is squared, so it grows much faster.
"DNE" again. Same trap as Q2. The limit exists and is a real number ($0$). DNE is for genuine non-existence, not for "the limit is $0$ or $\infty$".
Watch the limit direction. Both Q2 and Q3 are stated as $x \to \infty$. The same expression as $x \to -\infty$ can give different answers. For our Q3, the answer is the same ($0$) because both numerator and denominator go to $+\infty$. But for odd-degree numerators, signs can flip.

6 · Check your understanding

Practice

Use the degree-comparison rule to predict each limit, then check the answer:

$\displaystyle\lim_{x \to \infty} \frac{3x^2 + 1}{x^4 - 5}$ → degrees 2 vs 4, bottom wins
$\displaystyle\lim_{x \to \infty} \frac{7x^5 - x}{2x^5 + 3x^2}$ → degrees 5 vs 5, ratio of leading coefficients
$\displaystyle\lim_{x \to \infty} \frac{x^3 - 1}{x^2 + 1}$ → degrees 3 vs 2, top wins

Show answers

$0$ — denominator degree 4 > numerator degree 2, function dies off. Dominant behavior: $3x^2 / x^4 = 3/x^2 \to 0$.
$7/2$ — equal degrees, ratio of leading coefficients.
$\infty$ — numerator wins (degree 3 > 2). Same shape as Q2.

7 · What's next

You now have the three core cases of rational function limits at infinity. The next pretest problem might be one of:

Limits of the indeterminate form $\frac{0}{0}$ (where the degree rule doesn't apply and you need to factor) — this is the real work of the limits unit
One-sided limits and continuity
Limits with radicals or trig — conjugates, squeeze theorem
L'Hôpital's rule — the heavy artillery for hard $0/0$ and $\infty/\infty$ forms

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

← Q2: Limit at infinity (top wins) Q4: TBD → ← All Introduction (Pretest) questions

Lesson built by Aida for TK · Notre Dame MSDS