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

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AAI ATC Junior Executive 21 Feb 2023 Shift 2 Official Paper

Answer 1

Given:

\(\displaystyle \lim _{x \rightarrow ∞}\left(\frac{2 x-1}{2 x+3}\right)^{\frac{x+1}{2}} \)

Concept:

When 1∞ Case becomes then, we apply the concept

\(\lim_{x \rightarrow a}f^g=e^{\lim_{x \rightarrow a}g(f-1)}\)

Calculation:

\(\displaystyle \lim _{x \rightarrow ∞}\left(\frac{2 x-1}{2 x+3}\right)^{\frac{x+1}{2}} \)

\(=\displaystyle \lim _{x \rightarrow ∞}\left(\frac{2 -\frac{1}{x}}{2 +\frac{3}{x}}\right)^{\frac{x+1}{2}} \ \)

This is 1^∞ case then, we apply \(\lim_{x \rightarrow a}f^g=e^{\lim_{x \rightarrow a}g(f-1)}\)

\(=\displaystyle e^{\lim _{x \rightarrow ∞}\left({\frac{x+1}{2}}\right)\left(\frac{2 x-1}{2 x+3}-1\right)} \ \)

\(=\displaystyle e^{\lim _{x \rightarrow ∞}\left({\frac{x+1}{2}}\right)\left(\frac{-4}{2 x+3}\right)} \ \)

\(=\displaystyle e^{\lim _{x \rightarrow ∞}\left(\frac{-2x(+\frac{1}{x})}{2x(1+\frac{3}{2x})}\right)} \ \)

= e^-1 = 1/e

Hence option (4) is correct.

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