Find the limit of 1/(x sin x) − 1/x² as x → 0, with x in radians. Give the answer as a fraction in lowest terms.

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A difference of large terms

CalculusHard

Find the limit of $\frac{1}{x sin x} - \frac{1}{x ^{2}}$ as $x \to 0$ , with $x$ in radians. Give the answer as a fraction in lowest terms.

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Hint

Combine into one fraction and expand $sin x$ far enough to see the leading term.

Solution

Combine: $\frac{1}{x sin x} - \frac{1}{x ^{2}} = \frac{x - sin x}{x ^{2} sin x}$ .
From $sin x = x - \frac{x ^{3}}{6} + \dots$ the numerator is $\frac{x ^{3}}{6} + \dots$ .
The denominator is $x^{2} sin x = x^{3} + \dots$ .
The limit is therefore $\frac{\frac{1}{6}}{1} = \frac{1}{6}$ .

Answer: $\frac{1}{6}$