Limit

Calculus Level 4

lim ⁡ x → 0 ( 1 x sin ⁡ x − 1 tan ⁡ 2 x ) = a b \displaystyle\lim_{x \rightarrow 0} \left( \dfrac{1}{x\sin x} - \dfrac{1}{\tan^2 x}\right) = \dfrac{a}{b}

Given that a a and b b are positive integers where gcd ⁡ ( a , b ) = 1 \gcd (a,b)=1 , find the value of a + b a+b .

Inspiration


The answer is 11.

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Chan Lye Lee by Chan Lye Lee , 44, Singapore

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2 solutions

Otto Bretscher
Nov 21, 2015

lim ⁡ x → 0 ( ( sin ⁡ x ) / x − cos ⁡ 2 x sin ⁡ 2 x ) = lim ⁡ x → 0 ( ( sin ⁡ x ) / x − cos ⁡ 2 x x 2 ) = lim ⁡ x → 0 ( 1 − x 2 / 6 + o ( x 2 ) − ( 1 − x 2 + o ( x 2 ) ) x 2 ) = 5 6 \lim_{x \to 0}\left(\frac{(\sin x)/x - \cos^2 x}{\sin^2 x}\right)=\lim_{x \to 0}\left(\frac{(\sin x)/x - \cos^2 x}{x^2}\right)=\lim_{x \to 0}\left(\frac{1-x^2/6+o(x^2)-(1-x^2+o(x^2))}{x^2}\right)=\frac{5}{6} so that a + b = 11 a+b=\boxed{11}

10 Helpful 1 Interesting 1 Brilliant 0 Confused

Moderator note:

Good approach using error terms.

This is similar to the typical L'hopital rule, where we differentiate the numerator twice to find the limit.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

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