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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 61 of 160

Marks: +1, -0

If

\lfloor x \rfloor

is the greatest integer function, then

\lim\limits_{x \to 3}\; \frac{ \left(3-|x|+ \sin\;|3-x|\right) \cos \lfloor 9-3 x \rfloor }{ \left|3-x\right| \lfloor 3 x-9 \rfloor }

0
1
2
-2

Solution:

Since,

x \to 3^{-}

, we have

x < 3

.

\therefore |x|=x \;\text{ and }\; |3-x|=3-x

As,

x \to 3^{-},\; 3x \to 9^{-}

, so

9-3x \to 0^{+}

Thus,

\lfloor 9-3x \rfloor = 0

for

x

slightly less than 3 and

\lfloor 3x-9 \rfloor = -1

for

x

slightly less than 3 .

\therefore \lim\limits_{x \to 3^{-}} \frac{ \left(3-x+ \sin\;(3-x)\right) \cos(0) }{ (3-x)(-1) }

\;\; = \lim\limits_{x \to 3^{-}} \frac{ 3-x+ \sin\;(3-x) }{ (3-x)(-1) }

Let

t = 3-x

.

As,

x \to 3^{-},\; t \to 0^{+}

So,

\;\; \lim\limits_{x \to 3^{-}} \frac{ 3-x+ \sin\;(3-x) }{ (3-x)(-1) }

\Rightarrow \;\; \lim\limits_{t \to 0^{+}} \frac{ t+ \sin\;t }{ -t }

\Rightarrow \;\; \lim\limits_{t \to 0^{+}} \left( -1 - \frac{ \sin\;t }{ t } \right)

\Rightarrow \;\; -1-1 \;\; \left[ \because \lim\limits_{t \to 0} \frac{ \sin\;t }{ t } = 1 \right]

\Rightarrow \;\; -2

\therefore \lim\limits_{x \to 3^{-}} \frac{ \left(3-|x|+ \sin\;|3-x|\right) \cos \lfloor 9-3x \rfloor }{ \left|3-x\right| \lfloor 3x-9 \rfloor }

\; = -2

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