Before the midterm

Letting an oscillating factor decide the limit

The habit: Bound the wave between −1 and 1; then the product lies between −|g| and |g|, where g is the factor in front, and both bounds tend to 0 when g does.

3 practice questions are built to catch it, in 3 topics.

Practise against it: 2 questions

What it looks like

A question from limits, with the answer this slip gives beside the right one.

Standard2 marks · 3 min

Evaluate the limit.

\lim_{x \to \infty} \frac{5x - \cos(2x)}{2x + 1}

With the slip: dne
Worked through: $\displaystyle \frac{5}{2}$

cos(2x) keeps oscillating, but it stays between −1 and 1 while x grows without bound, so cos(2x)/x tends to 0.

See the worked solution

Where it comes up

Limits1 question
Continuity1 question
L’Hôpital’s rule1 question

Practise against it

Each of these has an answer this slip gives. Work them on paper: if you fall into it, the verdict names it and the question comes back in review.

01
For which value of the constant $k$ is the function below continuous at $x = 0$ ?
ContinuityStandard
02
Evaluate $\displaystyle\lim_{x \to 0} \frac{x^{2}\sin(1/x)}{\sin x}$ , or decide that it does not exist.
L’Hôpital’s ruleStandard

Start the set