Let fx - lim n - nn x - n x - n-2 - x - n-n-n- x2 - n2 x2 - n2-4 -x2 - n2-n2x-n- for all x 0- Then

F(x) = lim _n →∞ (n^n (x + n) (x + n/2) ⋯ (x + n/n)/n! (x^2 + n^2) (x^2 + n^2/4) ⋯(x^2 + n^2/n^2))^x/n nbsp; Taking log on both sides, we get ln f(x) = lim ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of Limit

Let fx = lim ...

Question

$Let f (x) = lim n \to \infty {⎛ ⎜ ⎜ ⎝ \frac{n^{n} (x + n) (x + \frac{n}{2}) \dots (x + \frac{n}{n})}{n! (x^{2} + n^{2}) (x^{2} + \frac{n^{2}}{4}) \dots (x^{2} + \frac{n^{2}}{n^{2}})} ⎞ ⎟ ⎟ ⎠}^{\frac{x}{n}},$ for all $x > 0.$ Then

f (\frac{1}{2}) \geq f (1)

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f (\frac{1}{3}) \leq f (\frac{2}{3})

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f^{'} (2) \leq 0

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\frac{f^{'} (3)}{f (3)} \geq \frac{f^{'} (2)}{f (2)}

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Solution

The correct option is C $f^{'} (2) \leq 0$
$f (x) = lim n \to \infty {⎛ ⎜ ⎜ ⎝ \frac{n^{n} (x + n) (x + \frac{n}{2}) \dots (x + \frac{n}{n})}{n! (x^{2} + n^{2}) (x^{2} + \frac{n^{2}}{4}) \dots (x^{2} + \frac{n^{2}}{n^{2}})} ⎞ ⎟ ⎟ ⎠}^{\frac{x}{n}}$

Taking log on both sides, we get
$ln f (x) = lim n \to \infty \frac{x}{n} ⎡ ⎢ ⎢ ⎢ ⎣ ln (\frac{n^{n}}{n!}) + n \sum r = 1 ln ⎛ ⎜ ⎜ ⎜ ⎝ \frac{x + \frac{n}{r}}{x^{2} + {(\frac{n}{r})}^{2}} ⎞ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎦$

$\Rightarrow ln f (x) = lim n \to \infty \frac{x}{n} ⎡ ⎢ ⎢ ⎢ ⎣ n \sum r = 1 ln (\frac{n}{r}) + n \sum r = 1 ln ⎛ ⎜ ⎜ ⎜ ⎝ \frac{x + \frac{n}{r}}{x^{2} + {(\frac{n}{r})}^{2}} ⎞ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎦$

$\Rightarrow ln f (x) = lim n \to \infty \frac{x}{n} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ n \sum r = 1 ln (\frac{1}{r / n}) + n \sum r = 1 ln ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{x + \frac{1}{r / n}}{x^{2} + {(\frac{1}{r / n})}^{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
$\Rightarrow ln f (x) = x ⎡ ⎢ ⎣ 1 \int 0 ln (1 / t) d t + 1 \int 0 ln (\frac{x + 1 / t}{x^{2} + 1 / t^{2}}) d t ⎤ ⎥ ⎦$

$\Rightarrow ln f (x) = x ⎡ ⎢ ⎣ 1 \int 0 (- ln t + ln (t x + 1) - ln t - ln (t^{2} x^{2} + 1) + ln t^{2}) d t ⎤ ⎥ ⎦$
Let $t x = z$
$Then ln f (x) = x \int 0 ln (\frac{1 + z}{1 + z^{2}}) d z \dots (1)$
$\Rightarrow f (x) = e^{x \int 0 ln ⎛ ⎝ \frac{1 + z}{1 + z^{2}} ⎞ ⎠ d z}$
Differentiating $(1)$ w.r.t. $x$ , we get
$\frac{f^{'} (x)}{f (x)} = ln (\frac{1 + x}{1 + x^{2}})$

We know that $f (x) > 0 \forall x > 0$
For $f^{'} (x) > 0$
$ln (\frac{1 + x}{1 + x^{2}}) > 0 \Rightarrow \frac{1 + x}{1 + x^{2}} > 1 \Rightarrow x > x^{2} \Rightarrow 0 < x < 1$
$∴ f (x)$ is increasing $\forall x \in (0, 1)$
Option (1) is incorrect.
Option (2) is correct.

Option (3):
$f^{'} (x) = f (x) \times ln (\frac{1 + x}{1 + x^{2}}) \Rightarrow f^{'} (2) = f (2) \times ln (3 / 5) \leq 0$
Statement is correct.

Option (4):
$\frac{f^{'} (3)}{f (3)} = ln (2 / 5) \frac{f^{'} (2)}{f (2)} = ln (3 / 5)$
Statement is incorrect.

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Similar questions

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Let f(x)=limn→∞⎛⎜ ⎜⎝nn(x+n)(x+n2)⋯(x+nn)n!(x2+n2)(x2+n24)⋯(x2+n2n2)⎞⎟ ⎟⎠xn, for all x>0. Then

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