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

The Correct Option is C

To solve this problem, we need to analyze the behavior of the product \(f(x)g(x)\) as \(x \to 0\).

Step 1: Analyze \(f(x)\)
For \(f(x) = \sqrt{n}\) if \(x \in \left(\frac{1}{n+1}, \frac{1}{n}\right]\), as \(x \to 0\), \(n\) must increase because the intervals \(\left(\frac{1}{n+1}, \frac{1}{n}\right]\) get smaller and closer to 0. Thus, \(f(x)\) can assume increasingly larger values since \(\sqrt{n}\) tends to increase as \(n\) increases.

Step 2: Analyze \(g(x)\) 
We know from the condition \(\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt < g(x) < 2\sqrt{x}\) for \(x\in(0,1)\).

Step 3: Estimation of the lower bound of \(g(x)\)
To find the behavior of \(\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt\) as \(x \to 0\), we evaluate this integral's tendency.
Change of variables: Let \(t = x\), so \(dt \approx dx\), and consider that \(\frac{1-t}{t}\to 1\) as \(t\to x\). On estimation, this integral behaves like: \(\approx x^2\cdot\sqrt{\frac{1-x^2}{x^2}} \approx x\) for small \(x\).

Step 4: Evaluate \(\lim_{x \to 0} f(x)g(x)\)
As \(x\to0\), we substitute: \(g(x) < 2\sqrt{x}\) leads to the approximation \(f(x)g(x) \approx \sqrt{n}\cdot2\sqrt{x}\). As \(n\) increases such that \(\frac{1}{n}\) reaches \(x\), the term \(\sqrt{n}\approx\frac{1}{\sqrt{x}}\) cancels out the \(\sqrt{x}\), leading to: \(2\cdot\sqrt{n}\cdot\sqrt{x}\approx 2\cdot\sqrt{x}\cdot\frac{1}{\sqrt{x}}=2\).
Ultimately, \(\lim_{x\to0}f(x)g(x)=2\).

Answer 2

We need to solve a one-sided limit to find the answer, otherwise \(\lim\limits_{x→0^-}\) doesn't exist because it is not in the domain.
\(f(x)=\sqrt{(\frac{1}{x})-1}\)        where (.) = least integer function
\(\lim\limits_{x→0^+}\int\limits_{x^2}^x\sqrt{\frac{1-t}{t}}dt.\sqrt{(\frac{1}{x})-1}\le \lim\limits_{x→0^+}f(x).g(x)\le\lim\limits_{x→o^+}\sqrt{(\frac{1}{x}-1)}\times2\sqrt{x}\)
So, \(\lim\limits_{x→0^+}\sqrt{(\frac{1}{x})-1}\times2\sqrt{x}=\lim\limits_{x→0^+}2\sqrt{x}\sqrt{[\frac{1}{x}]}\ \ (\frac{1}{x}∉ Z)\)
\(=\lim\limits_{x→0^+}2\sqrt{x\left(\frac{1}{x}-\left\{\frac{1}{x}\right\}\right)}=2\)
\(=\lim\limits_{x→0^+}2\sqrt{x(\frac{1}{2})}=2;(\frac{1}{x}∉Z)\)
\(\lim\limits_{x→0^+}\int\limits_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt.\sqrt{\frac{1}{x}-\left\{\frac{1}{x}\right\}}=\frac{\int\limits_{x^2}^x\sqrt{\frac{1-t}{t}}dt.\sqrt{1-x\left\{\frac{1}{x}\right\}}}{\sqrt{x}}\)
\(\lim\limits_{x→0^+}\frac{\int\limits_{x^2}^x\sqrt{\frac{1-t}{t}dt}}{\sqrt{x}}=\lim\limits_{x→0^+}\frac{\sqrt{\frac{1-x}{x}}-2x\sqrt{\frac{1=x^2}{x^2}}}{\frac{1}{2\sqrt{x}}}\)
\(\lim\limits_{x→0^+}2\sqrt{1-x}-4\sqrt{x}.\sqrt{1-x^2}=2\)
In the same way, \(\frac{1}{4}\in Z\) is equal to 2.
So, the correct option is (C) : is equal to 2.