Limit

Calculus Level 3

lim x a ( a x ) tan π x 2 a = ? \large \lim_{x\to a} (a -x) \tan\frac{\pi x}{2a} = \,?

2 a π \frac{2a}{\pi} a π a\pi Limit does not exist a 2 π \frac{a}{2\pi}

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1 solution

Let u = a x u = a - x . Then π x 2 a = π ( a u ) 2 a = π 2 π u 2 a \dfrac{\pi x}{2a} = \dfrac{\pi(a - u)}{2a} = \dfrac{\pi}{2} - \dfrac{\pi u}{2a} and u 0 u \to 0 as x a x \to a .

The given limit can then be written as

lim u 0 u tan ( π 2 π u 2 a ) = lim u 0 u cot ( π u 2 a ) = lim u 0 2 a π × π u 2 a tan ( π u 2 a ) = 2 a π lim y 0 y tan ( y ) \displaystyle\lim_{u \to 0} u \tan\left(\dfrac{\pi}{2} - \dfrac{\pi u}{2a}\right) = \lim_{u \to 0} u \cot\left(\dfrac{\pi u}{2a}\right) = \lim_{u \to 0} \dfrac{\dfrac{2a}{\pi} \times \dfrac{\pi u}{2a}}{\tan\left(\dfrac{\pi u}{2a}\right)} = \dfrac{2a}{\pi} \lim_{y \to 0} \dfrac{y}{\tan(y)} ,

where y = π u 2 a 0 y = \dfrac{\pi u}{2a} \to 0 as u 0 u \to 0 . Now lim y 0 tan ( y ) y = 1 \displaystyle \lim_{y \to 0} \dfrac{\tan(y)}{y} = 1 , so the desired limit is just 2 a π \boxed{\dfrac{2a}{\pi}} .

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