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# For JEE aspirants, a popular question …

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<font face = "Times New Roman">For JEE aspirants, a popular question of quest:
Find the greatest non-negative value of 'a' for which ,
$$\displaystyle\lim_{x \to 1}\left[\dfrac{a + sin(x-1) - ax}{x + sin(x-1) - 1}\right]^{\dfrac{1-x}{1-\sqrt{x}} = \dfrac{1}{4}$$
<font face ="Times New Roman">Thanks, but to see correct question add course of jee main and check my correct question there.

Sanskriti Kumari 2 months, 1 week ago

We know, lim n 0 ^ sin(u) m =1 lim sin(- 1/x) u -> e (1/x) then I∞ 91 tx + 1/4 = y 94 lim y -> ∞ (sin y)/y which means y (Delta*in - ∞)/(∞) tim Since, range of sine function is [-1,1] lim as divided any by constant number in 0.

☮🎭Amiable Buddy🎭☮ 2 months, 1 week ago

This is the question:
$$\displaystyle \lim_{x \to 1}\left[\dfrac{a + sin(x-1) - ax}{x + sin(x-1) - 1}\right]^{\dfrac{1-x}{1-\sqrt{x}} = \dfrac{1}{4}$$

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