Q99 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

$\lim _{x \rightarrow \infty} \frac{[2 x-3]}{x}=$

  • A
    $0$
  • B
    $\infty$
  • C
    $-3$
  • Correct answer
    D
    $2$

Solution

$\begin{aligned} & \text { We know, } x-1 \lt [x] \leq x \\ & \Rightarrow 2 x-4 \lt [2 x-3] \leq 2 x-3 \\ & \Rightarrow 2-\frac{4}{x} \lt \frac{[2 x-3]}{x} \leq \frac{2-3}{x} \\ & \lim _{x \rightarrow \infty}\left(2-\frac{4}{x}\right) \lt \lim _{x \rightarrow \infty} \frac{[2 x-3]}{x} \leq \lim _{x \rightarrow \infty}\left(2-\frac{3}{x}\right) \\ & \Rightarrow 2 \lt \lim _{x \rightarrow \infty} \frac{[2 x-3]}{x} \leq 2 . \text { Hence, } \lim _{x \rightarrow \infty} \frac{[2 x-3]}{x}=2\end{aligned}$

Hence, the correct answer is option: 'D'.

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