Do I need to use the epsilon-delta definition in MATH 100?

Rarely. Questions ask you to evaluate limits and reason about one-sided and infinite behaviour; a formal epsilon-delta proof is not the usual expectation in a first course.

When can I use L’Hôpital’s rule?

Practise limits by algebra first: factor, rationalise, or divide by the highest power, because a question can ask for a limit without the rule. The rule itself applies only to the forms 0/0 and infinity over infinity, and only when the limit of the quotient of the derivatives exists or is infinite; any other indeterminate form has to be rewritten as one of these first.

How much working should I show for a limit question?

Enough that each step follows from the one above. Naming the indeterminate form, showing the rewrite that removes it, and then substituting is normally the full expectation for a two or three mark question.

MATH 100 limits practice

A limit describes where a function is heading as its input approaches some value, which is not always where the function arrives, and sometimes the function never arrives at all. Almost everything that follows in MATH 100 is defined in terms of one, so the fluency you build here is spent for the rest of the term.

Most limit questions on a first-term paper fall into four shapes: substitution works immediately, substitution gives 0/0 and the expression needs rewriting, the two sides disagree, or the variable runs off to infinity. Recognising which shape you are looking at is most of the work; the algebra afterwards is usually short.

The questions below cover all four, from a factor-and-cancel warm-up to a bounded factor trapped against zero and a limit with an unknown constant.

Start practising Browse all 30 questions

In this topic

Taught: Before the midterm
Questions: 30
Total marks: 73
Suggested time: 120 min
Difficulty: Introductory to Capstone
Show the 9 hardest

Key results

One-sided limits: $\lim_{x \to a} f(x) = L \iff \lim_{x \to a^{-}} f(x) = L = \lim_{x \to a^{+}} f(x)$
Limit laws: $\lim_{x \to a}\bigl(f \pm g\bigr) = F \pm G, \qquad \lim_{x \to a} fg = FG, \qquad \lim_{x \to a}\frac{f}{g} = \frac{F}{G}$
When $\lim_{x \to a} f(x) = F$ and $\lim_{x \to a} g(x) = G$ both exist as real numbers, and for the quotient $G \neq 0$ .
Nonzero over zero: $\lim_{x \to a} f(x) = F \neq 0,\ \ \lim_{x \to a} g(x) = 0 \implies \lim_{x \to a}\left|\frac{f(x)}{g(x)}\right| = \infty$
With $g(x) \neq 0$ near $a$ ; the sign of $f/g$ on each side decides $+\infty$ or $-\infty$ , the two-sided limit is infinite only when both sides agree, and $F = 0$ as well is the form $0/0$ , which needs rewriting first.
The squeeze theorem: $g(x) \le f(x) \le h(x) \ \text{ and } \ \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \implies \lim_{x \to a} f(x) = L$
When the inequalities hold for every $x$ near $a$ , except possibly at $a$ itself; the same holds as $x \to \pm\infty$ when they hold for all large $|x|$ .
The sine limit: $\lim_{x \to 0}\frac{\sin x}{x} = 1$
With $x$ in radians.
Limits at infinity: $\lim_{x \to \pm\infty}\frac{1}{x^{n}} = 0, \qquad \lim_{x \to -\infty} e^{x} = 0, \qquad \sqrt{x^{2}} = |x|$
For any positive integer $n$ ; the last holds for every $x$ , so $\sqrt{x^{2}} = -x$ when $x < 0$ .

Read the theoryUBC Math 100 §2.1 Quick review of limitsUBC Math 100 §2.1.1 Calculating limits with limit lawsUBC Math 100 §2.1.2 Limits at infinityUBC Math 100 §2.2 AsymptotesCLP-1 §1.3 The limit of a functionCLP-1 §1.4 Calculating limits with limit lawsCLP-1 §1.5 Limits at infinity

Worked example

Standard2 marks · 3 min

Evaluate the limit.

\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}

Strategy

The indeterminate form comes from a difference of square roots. Multiplying by the conjugate turns that difference into a difference of squares, which cancels the troublesome factor.

Working

$\lim_{x \to 0}\bigl(\sqrt{x + 4} - 2\bigr) = 0, \qquad \lim_{x \to 0} x = 0$
Substitution gives 0/0, so the expression has to be rewritten before the limit can be read off.
$\frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2}$
Multiply by the conjugate over itself. This is multiplication by 1, so the value is unchanged.
$= \frac{(x + 4) - 4}{x\bigl(\sqrt{x+4} + 2\bigr)} = \frac{x}{x\bigl(\sqrt{x+4} + 2\bigr)}$
The numerator collapses by the difference-of-squares identity. Leave the denominator factored.
$= \frac{1}{\sqrt{x+4} + 2} \quad \text{for } x \neq 0$
Cancel the shared factor of x. The indeterminate form is gone.
$\lim_{x \to 0} \frac{1}{\sqrt{x+4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$
Substitute x = 0 into the continuous expression that remains.

Answer

\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} = \frac{1}{4}

Check

At x = 0.01 the quotient is (√4.01 − 2)/0.01 ≈ 0.2498, close to 1/4.

Where this goes wrong

Expanding the root term by term as √x + √4 and cancelling. The square root of a sum is not the sum of the square roots, so that step is not algebra you can do.

On the paper

Say what substitution gives, $0/0$ or a nonzero number over $0$ , before you rewrite. Keep $\lim$ on every line until the last substitution, and give the sign from each side for an infinite limit.

Leans on

Rationalising with a conjugate, from Functions, graphs and inverses

Your turn

Challenging3 marks · 5 min

Evaluate the limit.

\lim_{x \to 2} \frac{\sqrt{5x + 6} - \sqrt{x + 14}}{x - 2}

Answer this question

What you should be able to do

Decide whether substitution is enough
For a function continuous at the point, the limit is just the value. Knowing when you are allowed to stop there saves time you will want later in the paper.
Resolve the indeterminate form 0/0
Factor and cancel, multiply by a conjugate, or combine fractions over a common denominator. Each technique targets a different reason the zero appeared.
Work with one-sided limits
Evaluate from the left and the right separately, and state that the two-sided limit exists only when they agree. Piecewise functions and vertical asymptotes both depend on this.
Handle limits at infinity
Divide the top and the bottom by the highest power of $x$ in the denominator and let every reciprocal term vanish. Under a square root, $\sqrt{x^{2}} = |x|$ , which is $-x$ as $x \to -\infty$ . For rational functions this reduces to comparing leading terms, sign included.
Trap a bounded factor
A factor such as $\sin(1/x)$ oscillates without settling, but it stays between −1 and 1. Multiply that bound by a factor that tends to zero and is never negative, such as $x^{2}$ or $|x|$ , and the product is trapped between two quantities that both tend to zero.
State continuity in terms of limits
A function is continuous at a point when the limit exists, the value exists, and the two are equal. Questions asking you to choose a constant are testing exactly this definition.

Where marks are usually lost

Reading 0/0 as “the limit does not exist”
An indeterminate form is a signal that the expression has not been simplified, not a verdict. The expression is undefined at the point; the limit near it may still be an ordinary number.
Splitting a root across a sum
The square root of a sum is not the sum of the square roots. When a root is causing the trouble, the tool is the conjugate, not distribution.
Losing a sign in a limit at infinity
In an expression such as $\dfrac{3x^{2} - 5x}{7 - 2x^{2}}$ the leading coefficient of the denominator is −2, not 2. Rewriting the denominator in descending order before comparing prevents this.
Applying $\sin(u)/u$ when the arguments do not match
The standard limit equals 1 only when the argument of the sine and the denominator are identical and tend to 0. For $\lim_{x \to 0}\dfrac{\sin(3x)}{5x}$ the answer is $\dfrac{3}{5}$ , reached by forcing the match first.
Answering an unsigned infinity at a vertical asymptote
When the two sides run to $+\infty$ and $-\infty$ the two-sided limit does not exist, and the marks are for saying which side does which. A single unsigned answer cannot be right.

Practice questions

5 of the 30, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Browse all 30 questions

Common questions

Do I need to use the epsilon-delta definition in MATH 100?: Rarely. Questions ask you to evaluate limits and reason about one-sided and infinite behaviour; a formal epsilon-delta proof is not the usual expectation in a first course.
When can I use L’Hôpital’s rule?: Practise limits by algebra first: factor, rationalise, or divide by the highest power, because a question can ask for a limit without the rule. The rule itself applies only to the forms 0/0 and infinity over infinity, and only when the limit of the quotient of the derivatives exists or is infinite; any other indeterminate form has to be rewritten as one of these first.
How much working should I show for a limit question?: Enough that each step follows from the one above. Naming the indeterminate form, showing the rewrite that removes it, and then substituting is normally the full expectation for a two or three mark question.

Needs
Functions, graphs and inverses
Domain, composition, transformations, inverses and the trigonometry the course assumes.
19 questions
Leads to
The definition of the derivative
Secants and tangents, average and instantaneous velocity, the difference quotient and differentiability.
27 questions
Leads to
Curve sketching
Increase, concavity, inflection points, asymptotes and full sketches.
29 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

What you should be able to do

Decide whether substitution is enough

Resolve the indeterminate form 0/0

Work with one-sided limits

Handle limits at infinity

Trap a bounded factor

State continuity in terms of limits

Where marks are usually lost

Reading 0/0 as “the limit does not exist”

Splitting a root across a sum

Losing a sign in a limit at infinity

Applying sin⁡(u)/u\sin(u)/usin(u)/u when the arguments do not match

Answering an unsigned infinity at a vertical asymptote

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Common questions

Related topics

Functions, graphs and inverses

The definition of the derivative

Curve sketching

Applying $\sin(u)/u$ when the arguments do not match