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Limits: Gif

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

Question
lim⁡x→0([100sin⁡xx]+[100tan⁡xx])\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
A199199correct
BDoes not exist
C201201
D200200
Solution
Step 1: For x∈(0,π/2)x\in(0,\pi/2): sin⁡x<x<tan⁡x\sin x<x<\tan x, so sin⁡xx<1\dfrac{\sin x}{x}<1 and tan⁡xx>1\dfrac{\tan x}{x}>1. Step 2: As x→0x\to 0: sin⁡xx→1−\dfrac{\sin x}{x}\to 1^{-} (slightly less than 11) and tan⁡xx→1+\dfrac{\tan x}{x}\to 1^{+} (slightly greater than 11). Step 3: Therefore: - 100sin⁡xx\dfrac{100\sin x}{x} approaches 100−100^{-} (just below 100100), so [100sin⁡xx]=99\left[\dfrac{100\sin x}{x}\right]=99. - 100tan⁡xx\dfrac{100\tan x}{x} approaches 100+100^{+} (just above 100100), so [100tan⁡xx]=100\left[\dfrac{100\tan x}{x}\right]=100. Step 4: Sum:
99+100=19999+100=199
 Correct answer: (1) 199199
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