Before Quiz 2

Concluding there is no limit because f′/g′ has none

The habit: When f′/g′ has no limit, l’Hôpital’s rule says nothing about f/g, so settle it another way, such as dividing through by the fastest-growing term.

3 practice questions are built to catch it, in 1 topic.

Practise against it: 2 questions

What it looks like

A question from L’Hôpital’s rule, with the answer this slip gives beside the right one.

Standard3 marks · 4 min

Evaluate $\displaystyle\lim_{x \to 0} \frac{x^{2}\sin(1/x)}{\sin x}$ , or decide that it does not exist.

With the slip: The limit does not exist, because l’Hôpital’s rule gives $\dfrac{2x\sin(1/x) - \cos(1/x)}{\cos x}$ , which has no limit as $x \to 0$ .
Worked through: $\displaystyle 0$

The rule says nothing when f′/g′ has no limit. Split the quotient instead, as (x/sin x) · x sin(1/x), and take each factor’s limit.

See the worked solution

Where it comes up

L’Hôpital’s rule3 questions

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
Evaluate $\displaystyle\lim_{x \to \infty} \frac{x + \sin x}{x}$ , or decide that it does not exist.
L’Hôpital’s ruleChallenging
02
Two students start on the limit below. Student P applies l’Hôpital’s rule, and student Q divides the top and the bottom by $3^{x}$ . Which statement is right? $\displaystyle \lim_{x \to \infty} \frac{4\cdot 3^{x} + 2^{x}}{3^{x} + 5\cdot 2^{x}}$
L’Hôpital’s ruleChallenging

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