​$$\text { The value of } \lim _{x \rightarrow \infty}\left(\frac{2 x-1}{2 x+3}\right)^{\frac{x+1}{2}} \text { is: }$$​​

Correct option is D

​$$\text{We are given:} \\[6pt]\lim_{x \to \infty} \left( \frac{2x - 1}{2x + 3} \right)^{\frac{x+1}{2}} \\[10pt]\text{Step 1: Simplify the expression inside the limit:} \\[6pt]\frac{2x - 1}{2x + 3} = \frac{2x(1 - \frac{1}{2x})}{2x(1 + \frac{3}{2x})} = \frac{1 - \frac{1}{2x}}{1 + \frac{3}{2x}} \\[10pt]\text{As } x \to \infty, \ \frac{1}{2x} \to 0, \text{ so:} \\[6pt]\frac{2x - 1}{2x + 3} \to 1 \ \text{from below} \\[10pt]\text{Let } f(x) = \left( \frac{2x - 1}{2x + 3} \right)^{\frac{x+1}{2}} \\[6pt]\text{Take natural log:} \\[6pt]\ln f(x) = \frac{x+1}{2} \cdot \ln \left( \frac{2x - 1}{2x + 3} \right) \\[10pt]\text{Use expansion: } \ln(1 - \epsilon) \approx -\epsilon \text{ for small } \epsilon \\[6pt]\frac{2x - 1}{2x + 3} = 1 - \frac{4}{2x + 3} \\[6pt]\Rightarrow \ln \left( \frac{2x - 1}{2x + 3} \right) \approx -\frac{4}{2x + 3} \\[10pt]\Rightarrow \ln f(x) \approx \frac{x+1}{2} \cdot \left( -\frac{4}{2x + 3} \right) \\[10pt]\text{As } x \to \infty: \\[6pt]\ln f(x) \to \frac{x}{2} \cdot \left( -\frac{4}{2x} \right) = -1 \Rightarrow f(x) \to e^{-1} \\[10pt]\text{Final Answer:} \quad {\dfrac{1}{e}}$$

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