OMPT drill — Calculus

Limits

Limits at infinity and around excluded points, mostly via algebraic simplification. On the OMPT this stays computational — no epsilon-delta in sight. Say the rule you are using out loud before you differentiate or integrate — it sounds silly and it works.

Tested inOMPT-D

Lesson

A limit asks what value a function is heading toward, which is a different question from what value it takes. $\frac{x ^{2} - 9}{x - 3}$ is undefined at $x = 3$ , plugging in gives $\frac{0}{0}$ , yet as $x$ creeps toward 3 from either side, the outputs creep toward 6. We write $lim_{x \to 3} \frac{x ^{2} - 9}{x - 3} = 6$ : the destination exists even though the function never stands on it.

On the OMPT this topic stays computational, and that is good news. No epsilon-delta arguments, no pathological cases: just two situations, each with a standard fix. Situation one: substituting gives $\frac{0}{0}$ , which means a common factor is hiding in top and bottom; factor, cancel, substitute again. Situation two: $x$ runs off to infinity, and the question becomes a contest between growth rates that the highest powers always win. Learn the two fixes and every limit on the test yields to one of them.

Problem 1Numeric answer

Evaluate $x \to 3 lim \frac{x ^{2} - 9}{x - 3}$ .

Show the worked solution

Answer: 6

Substituting $x = 3$ gives $\frac{0}{0}$ , so I factor: $\frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3$ for $x \neq = 3$ . The limit only cares about values near 3, so the answer is $3 + 3 = 6$ .

Problem 2Multiple choice

Evaluate $x \to \infty lim \frac{5 x ^{2} - 3 x}{2 x ^{2} + 7}$ .

A $0$
B $\frac{5}{2}$
C $\frac{5}{7}$
DThe limit does not exist

Show the worked solution

Answer: B — $\frac{5}{2}$

Dividing top and bottom by $x^{2}$ gives $\frac{5 - 3/ x}{2 + 7/ x ^{2}}$ . As $x \to \infty$ the small terms vanish, leaving $\frac{5}{2}$ . With equal degrees, the limit is always the ratio of leading coefficients.

Problem 3Numeric answer

Evaluate $x \to 0 lim \frac{sin 4 x}{x}$ .

Show the worked solution

Answer: 4

I match the argument of the sine with the denominator: $\frac{s i n 4 x}{x} = 4 \cdot \frac{s i n 4 x}{4 x}$ . Since $\frac{s i n u}{u} \to 1$ as $u \to 0$ , the limit is $4 \cdot 1 = 4$ .

Problem 4Multiple choice

Which of the following limits does NOT exist?

A $x \to 0 lim \frac{∣ x ∣}{x}$
B $x \to 0 lim x sin \frac{1}{x}$
C $x \to 2 lim \frac{x ^{2} - 4}{x - 2}$
D $x \to 0 lim \frac{sin x}{x}$

Show the worked solution

Answer: A — $x \to 0 lim \frac{∣ x ∣}{x}$

For $\frac{∣ x ∣}{x}$ , the value is $1$ from the right and $- 1$ from the left; the one-sided limits disagree, so the limit fails to exist. The others all converge: the squeeze theorem gives 0 for the second, factoring gives 4 for the third, and the standard limit gives 1 for the fourth.

Problem 5Spot the error

A student evaluates $x \to 1 lim \frac{x ^{2} + 1}{x + 1}$ . Step 1: this looks like a limit of a quotient, so apply l'Hôpital's rule. Step 2: differentiate top and bottom to get $\frac{2 x}{1}$ . Step 3: substitute $x = 1$ to get 2. Which step contains the error?

Show the worked solution

Answer: Step 1

The error is the very first move. L'H\^opital's rule only applies to the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , but here direct substitution works fine: $\frac{1 ^{2} + 1}{1 + 1} = \frac{2}{2} = 1$ . Steps 2 and 3 execute the (inapplicable) rule correctly, which is why the wrong answer 2 looks so convincing.

Problem 6Numeric answer

The function $f (x) = \frac{x ^{2} - 4}{x - 2}$ for $x \neq = 2$ , with $f (2) = a$ , is continuous at $x = 2$ . Find $a$ .

Show the worked solution

Answer: 4

Continuity at $x = 2$ demands $a = lim_{x \to 2} f (x)$ . Factoring, $\frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2$ near 2, so the limit is 4 and therefore $a = 4$ .