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Let $f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{n^n}\left( {x + n} \right)\left( {x + \frac{n}{2}} \right)..\left( {x + \frac{n}{n}} \right)}}{{n!\left( {{x^2} + {n^2}} \right)\left( {{x^2} + \frac{{{n^2}}}{4}} \right)..\left( {{x^2} + \frac{{{n^2}}}{{{n^2}}}} \right)}}} \right)^{\frac{x}{n}}}$, for all $x>0$ then

(A) $f(\frac{1}{2})\ge f(1)$

(B) $f(\frac{1}{3})\le f(\frac{2}{3})$

(C) $f'(2)\le 0$

(D) $\frac{f'(3)}{f(3)}\ge \frac{f'(2)}{f(2)}$

The official answer is (B) and (C)

My approach is as follow $\ell nf\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\left( {x + n} \right)\left( {x + \frac{n}{2}} \right)..\left( {x + \frac{n}{n}} \right)} \right) - \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\frac{{n!}}{{{n^n}}}\left( {{x^2} + {n^2}} \right)\left( {{x^2} + \frac{{{n^2}}}{4}} \right)..\left( {{x^2} + \frac{{{n^2}}}{{{n^2}}}} \right)} \right)$

$\ell nf\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\left( {x + n} \right)\left( {x + \frac{n}{2}} \right)..\left( {x + \frac{n}{n}} \right)} \right) - \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\frac{{n!}}{{{n^n}}}} \right) - \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\left( {{x^2} + {n^2}} \right)\left( {{x^2} + \frac{{{n^2}}}{4}} \right)..\left( {{x^2} + \frac{{{n^2}}}{{{n^2}}}} \right)} \right)$

$\ell nf\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\left( {x + \frac{1}{{\frac{1}{n}}}} \right)\left( {x + \frac{1}{{\frac{2}{n}}}} \right)..\left( {x + \frac{1}{{\frac{n}{n}}}} \right)} \right) - \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\frac{{n!}}{{{n^n}}}} \right) - \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\left( {{x^2} + \frac{1}{{\frac{{{1^2}}}{{{n^2}}}}}} \right)\left( {{x^2} + \frac{1}{{\frac{{{2^2}}}{{{n^2}}}}}} \right)..\left( {{x^2} + \frac{1}{{\frac{{{n^2}}}{{{n^2}}}}}} \right)} \right)$

$\ell nf\left( x \right) = x\int\limits_0^1 {\ell n\left( {x + \frac{1}{t}} \right)dt} - \mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\frac{{n!}}{{{n^n}}}} \right) - x\int\limits_0^1 {\ell n\left( {{x^2} + \frac{1}{{{t^2}}}} \right)dt} $

Not able to proceed from here

Can i use $\mathop {\lim }\limits_{n \to \infty } \frac{x}{n}\ell n\left( {\frac{{n!}}{{{n^n}}}} \right) = x\int\limits_0^1 {\ell ntdt} $

Samar Imam Zaidi
  • 8,800

1 Answers1

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HINT: ${\frac{{n!}}{{{n^n}}}}=\frac1n\frac2n...\frac nn$. You would be able to convert limit into integration. Then take LCM for $x+\frac1t$ and $x^2+\frac1{t^2}$, and put $xt=y$, your limits of integration would change to zero to $x$. Then, take derivative of the whole equation without solving integration. Maxima would emerge at $x=1$. So, Option $A$ is false. Between $0$ to $1$, $f'(x)$ would come out to be positive i.e. $f(x)$ is increasing function in $(0,1)$. So, option $B$ is correct. $f'(x)$ would come out to be negative when $x\gt 1$. So, Option $C$ is correct. It would be clear early on that option $D$ is false. Are you able to solve it now? Any doubt?

aarbee
  • 8,246