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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\lim_{x\to 0}\left(\left[\dfrac{100\sin x}{x}\right]+\left[\dfrac{100\tan x}{x}\right]\right)

, where

[\cdot]

is GIF

199

correct

BDoes not exist

201

200

Solution

Step 1: For

x\in(0,\pi/2)

\sin x<x<\tan x

, so

\dfrac{\sin x}{x}<1

and

\dfrac{\tan x}{x}>1

. Step 2: As

x\to 0

\dfrac{\sin x}{x}\to 1^{-}

(slightly less than

1

) and

\dfrac{\tan x}{x}\to 1^{+}

(slightly greater than

1

). Step 3: Therefore: -

\dfrac{100\sin x}{x}

approaches

100^{-}

(just below

100

), so

\left[\dfrac{100\sin x}{x}\right]=99

. -

\dfrac{100\tan x}{x}

approaches

100^{+}

(just above

100

), so

\left[\dfrac{100\tan x}{x}\right]=100

. Step 4: Sum:

99+100=199

Correct answer: (1)

199

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