L’Hôpital’s rule, question 16: UBC MATH 100 practice

Question 16 of 27Standard3 marks · 4 minTheory UBC Math 100 §6.1CLP-1 §3.7Key results

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

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 $x \to 0$ .
$0$
$1$
The limit does not exist, because $\sin(1/x)$ oscillates between $-1$ and $1$ as $x \to 0$ .

How sure are you?

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Theory UBC Math 100 §6.1 CLP-1 §3.7Key results

Worked solution

Strategy

The form is 0/0, but the quotient of derivatives has no limit, so l’Hôpital’s rule is silent. Split the quotient into two factors whose limits are known.

Working

$\frac{\bigl(x^{2}\sin(1/x)\bigr)'}{(\sin x)'} = \frac{2x\sin(1/x) - \cos(1/x)}{\cos x}$
Product rule and chain rule: the derivative of sin(1/x) is cos(1/x) · (−1/x²), and the x² cancels.
$2x\sin(1/x) \to 0, \qquad \cos x \to 1, \qquad \cos(1/x) \text{ has no limit}$
Because cos(1/x) keeps swinging between −1 and 1, f′/g′ has no limit as x → 0, and the rule gives no information.
$\frac{x^{2}\sin(1/x)}{\sin x} = \frac{x}{\sin x}\cdot x\sin\!\left(\frac{1}{x}\right), \qquad \left|x\sin\!\left(\frac{1}{x}\right)\right| \le |x| \to 0$
x / sin x → 1, and a bounded factor times x tends to 0.
$\lim_{x \to 0} \frac{x^{2}\sin(1/x)}{\sin x} = 1\cdot 0 = 0$
The limit exists, although the rule could not find it.

Answer

0

Check

At x = 0.01 the quotient is about −0.0051, and in size it is at most x²/|sin x|, which tends to 0.

Where this goes wrong

Concluding that the limit does not exist because f′/g′ has none. The rule works in one direction only: if f′/g′ has a limit, f/g has the same one; if it has none, the rule is silent and another method is needed.

The idea behind it

The L’Hôpital’s rule lesson, a figure to move, in a few steps

Leans on

Limits of products, from Limits

Try a similar question

Evaluate

\displaystyle\lim_{x \to \infty} \frac{x + \sin x}{x}

, or decide that it does not exist.

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L’Hôpital’s rule practice question 16