Let f : (0, 1) → R be the function defined as f(x)=n if x∈1n+1,1n where n ∈ N. Let g : (0, 1) → R be a function such that ∫x2x 1−ttdt<g(x)<2x for all x ∈ (0, 1). Then limx→0 f(x)g(x)

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Let f : (0, 1) → R be the function defined as $f (x) = \sqrt{n} if x \in [\frac{1}{n + 1}, \frac{1}{n})$ where n ∈ N. Let g : (0, 1) → R be a
function such that $\int_{x^{2}}^{x} \sqrt{\frac{1 - t}{t}} dt < g (x) < 2 \sqrt{x}$ for all x ∈ (0, 1). Then $lim_{x \to 0} f (x) g (x)$

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Is equal to 2

Is equal to 1

Is equal to 3

Does not exist

answer is C.

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Detailed Solution

$f (x) = \sqrt{n}, x \in [\frac{1}{n + 1}, \frac{1}{n}) \begin{array}{l} \frac{1}{n + 1} \leq x < \frac{1}{n} \Rightarrow n + 1 \geq \frac{1}{x} > n \\ \Rightarrow n \geq \frac{1}{x} - 1 and n < \frac{1}{x} \Rightarrow \sqrt{n} \geq \sqrt{\frac{1}{x} - 1} and \sqrt{n} < \frac{1}{\sqrt{x}} \\ lim_{x \to 0} f (x) g (x) = \\ \int_{x^{2}}^{x} \sqrt{\frac{1 - t}{t}} dt < g (x) < 2 \sqrt{x} \Rightarrow f (x) \int_{x^{2}}^{x} \sqrt{\frac{1 - t}{t}} dt < f (x) g (x) < f (x) (2 \sqrt{x}) \\ \Rightarrow (\int_{x^{2}}^{x} \sqrt{\frac{1 - t}{t} dt}) (\sqrt{\frac{1}{x} - 1}) < f (x) g (x) < \frac{1}{\sqrt{x}} (2 \sqrt{x}) \\ \Rightarrow lim_{x \to 0} (\int_{x^{2}}^{x} \sqrt{\frac{1 - t}{t}} dt) (\sqrt{\frac{1}{x} - 1}) < lim_{x \to 0} f (x) g (x) < lim_{x \to 0} 2 \\ \begin{array}{l} \Rightarrow lim_{x \to 0} [\frac{[\sqrt{x (1 - x)} - x \sqrt{1 - x^{2}} + \sin^{- 1} \sqrt{x} - \sin^{- 1} (x)] \sqrt{(1 - x)}}{\sqrt{x}}] < lim_{x \to 0} f (x) g (x) < 2 \\ \Rightarrow (1 - 0 + 1 - 0) < lim_{x \to 0} f (x) g (x) < 2 \\ \Rightarrow lim_{x \to 0} f (x) g (x) = 2 \end{array} \end{array}$