ₙ → ∞}\; {(2n(2n-1)(n+2)(n+1))¹/n}}{n}=

Correct Answer is : ∫01log⁡(1+x)dx₀¹ (1+x) dx0∫1log(1+x)dx

Correct Answer is : ∫01log⁡(1+x)dx₀¹ (1+x) dx0∫1log(1+x)dx

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TG EAMCET 4 May 2025 Shift 2 Paper

Question : 76 of 160

Marks: +1, -0

\lim\limits_{n \to \infty}\; \frac{(2n(2n-1)\dots(n+2)(n+1))^{1/n}}{n}=

Solution:

\text{ Let } \lim\limits_{n \to \infty} \left( \frac{(n+1)(n+2)\dots(2n)}{n^n} \right)^{\frac{1}{n}} = y

\Rightarrow \ln y = \lim\limits_{n \to \infty} \frac{1}{n} \left( \sum\limits_{r=1}^{n} \ln (n+r) - \sum\limits_{r=1}^{n} \ln (n) \right)

\Rightarrow \ln y = \lim\limits_{n \to \infty} \frac{1}{n} \left( \sum\limits_{r=1}^{n} \ln \left(1+\frac{r}{n}\right) \right)

\Rightarrow \ln y = \int\limits_{0}^{1} \ln (1+x) dx

\Rightarrow y = e^{0} \ln (1+x) dx

\text{ Let } P = \int\limits_{0}^{1} \ln (1+x) dx

\Rightarrow P = \left[ x \cdot \ln (1+x) \right]_{0}^{1} - \int\limits_{0}^{1} \left( \frac{x}{1+x} \right) dx

\Rightarrow P = \ln 2 - \int\limits_{0}^{1} \left( \frac{x}{1+x} \right) dx

\text{ Again, letl } +x = t

dx = dt

= \int\limits_{1}^{2} \left( \frac{t-1}{t} \right) dt = \int\limits_{1}^{2} \left( 1 - \frac{1}{t} \right) dt

= \left[ t - \ln t \right]_{1}^{2}

= 2 - \ln 2 - 1 = 1 - \ln 2

\therefore P = \ln 2 - (1 - \ln 2)

\Rightarrow P = \ln 2 - 1 + \ln 2

\Rightarrow P = 2 \ln 2 - 1

\Rightarrow P = \ln 4 - \ln e

\Rightarrow P = \ln \frac{4}{e}

\Rightarrow y = e \ln \frac{4}{e}

y = \frac{4}{e}

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