Question

Let f : (0,1) → R be the function defined as f(x) = √n if x ∈ [$\frac{1}{n+1},\frac{1}{n}$] 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 $\lim_{x\rightarrow0}f(x)g(x)$

• A. does NOT exist
• B. is equal to 1
• C. is equal to 2
• D. is equal to 3

Answer 1

Answer: (C)

is equal to 2

Answer 2

The correct option is (C).
$f(x)=\sqrt{\frac{1}{x}-1}$
$\lim_{x\rightarrow 0^+}\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt\sqrt{\frac{1}{x}-1}\times2\sqrt x$
$=\lim_{x\rightarrow 0^+}2\sqrt{x(\frac{1}{x}-{\frac{1}{x}})}=2$
$\lim_{x\rightarrow 0^+}2\sqrt {x(\frac{1}{x})}=2;(\frac{1}{x}\notin Z)$
$\lim_{x\rightarrow0^+}\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt.\sqrt{\frac{1}{x}-\frac{1}{x}}=\int_{x^2}^{x}\frac{1-t}{t}dt\sqrt{1-x}(\frac{1}{x})$
$\lim_{x\rightarrow}2\sqrt{1-x}.4\sqrt{x}.\sqrt{1-x^2}=2$
similarly for $\frac{1}{x}\in Z$ is equal to 2.