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EXERCISE 8.1 · Q21

Q.Discuss the continuity of the following function at the point indicated against it: f(x)=3−tan⁡xπ−3xf(x) = \dfrac{\sqrt3 - \tan x}{\pi - 3x}, x≠π3x \ne \dfrac{\pi}{3}, =34= \dfrac{3}{4}, for x=π3x = \dfrac{\pi}{3}, at x=π3x = \dfrac{\pi}{3}.

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f(x)=3−tan⁡xπ−3xf(x)=\dfrac{\sqrt3-\tan x}{\pi-3x} for x≠π/3x\ne\pi/3, f(π/3)=3/4f(\pi/3)=3/4.

Put x=π/3+tx=\pi/3+t, so π−3x=−3t\pi-3x=-3t, and t→0t\to0 as x→π/3x\to\pi/3. Using the tangent addition formula, tan⁡(π/3+t)=3+tan⁡t1−3tan⁡t\tan(\pi/3+t)=\dfrac{\sqrt3+\tan t}{1-\sqrt3\tan t}, so

3−tan⁡x=3−3+tan⁡t1−3tan⁡t=3(1−3tan⁡t)−(3+tan⁡t)1−3tan⁡t=−4tan⁡t1−3tan⁡t.\sqrt3-\tan x=\sqrt3-\frac{\sqrt3+\tan t}{1-\sqrt3\tan t}=\frac{\sqrt3(1-\sqrt3\tan t)-(\sqrt3+\tan t)}{1-\sqrt3\tan t}=\frac{-4\tan t}{1-\sqrt3\tan t}.

So

f(x)=−4tan⁡t/(1−3tan⁡t)−3t=4tan⁡t3t(1−3tan⁡t)  ⟶  43as t→0,f(x)=\frac{-4\tan t/(1-\sqrt3\tan t)}{-3t}=\frac{4\tan t}{3t(1-\sqrt3\tan t)}\;\longrightarrow\;\frac{4}{3}\quad\text{as } t\to0, …

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