L’Hôpital’s rule, question 13

Topic: L’Hôpital’s rule
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §6 CLP-1 §3.7Key results

Evaluate the limit.

\lim_{x \to 0^{+}} \left(\sin x\right)^{2/\log x}

Your answer

An exact expression, or a decimal to two decimal places.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §6 CLP-1 §3.7Key results

Worked solution

Strategy

A variable base raised to a variable power: take the logarithm to bring the exponent down, settle the resulting quotient with the rule, and exponentiate at the end.

Working

$\sin x \to 0^{+}, \qquad \frac{2}{\log x} \to 0^{-} \qquad \bigl(\text{form } 0^{0}\bigr)$
Indeterminate: the base pulls towards 0 and the exponent towards a zeroth power.
$\lim_{x \to 0^{+}} \log\Bigl((\sin x)^{2/\log x}\Bigr) = \lim_{x \to 0^{+}} \frac{2\log(\sin x)}{\log x} \qquad \left(\tfrac{-\infty}{-\infty}\right)$
Work with the logarithm of the expression: it turns the power into a product, written here as a quotient.
$\overset{\text{l'H}}{=} \lim_{x \to 0^{+}} \frac{2\cos x/\sin x}{1/x} = \lim_{x \to 0^{+}} 2\cos x\cdot\frac{x}{\sin x} = 2\cdot 1\cdot 1 = 2$
The derivative of log(sin x) is cos x / sin x by the chain rule; x / sin x → 1 is the standard limit.
$\lim_{x \to 0^{+}} (\sin x)^{2/\log x} = \lim_{x \to 0^{+}} e^{\log\left((\sin x)^{2/\log x}\right)} = e^{2} \approx 7.39$
The expression is e to the power of its own logarithm, and eᵘ is continuous, so the limit passes into the exponent.

Answer

e^{2}

Check

At x = 0.1 the power is about 7.400 and at x = 0.01 about 7.3891, closing in on e² ≈ 7.3891.

Where this goes wrong

Answering 0 because the base tends to 0, or 1 because the exponent tends to 0. In a 0⁰ form the two pull against each other, and here the logarithm shows the balance is e², neither 0 nor 1.

Leans on

Laws of logarithms, from Functions, graphs and inverses

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