 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/20/c/ii/solution

Near zero,

cos x = 1 - x^{2} /2 + O (x^{4})

and

ln (1 + v) = v + O (v^{2})

, so

ln cos x = - x^{2} /2 + O (x^{4})

. Replacing

x

3 x

similarly gives

ln cos 3 x = - 9 x^{2} /2 + O (x^{4})

. Therefore

x \to 0 lim \frac{ln cos x}{ln cos 3 x} = \frac{1}{9} .

For all sufficiently small nonzero

x

, both cosines are positive and the denominator logarithm is nonzero, so this quotient and its two-sided real limit are well-defined.

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