rrow R, \\\\ & f(x)=\sqrt{n}, x \in\left[\frac{1}{n+1}, \frac{1}{n}\right), n \in N\end{aligned}$

$g:(0,1) \rightarrow R$ where

$$ \int_{x^2}^x \sqrt{\frac{1-t}{t}} d t < g(x) < 2 \sqrt{x}, x \in(0,1) $$

Now (According to the question)

$$ \begin{aligned} & \lim _{x \rightarrow \infty} f(x) \cdot g(x) \\\\ & \Rightarrow \text { Put } x=\frac{1}{n} \\\\ & \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \end{aligned} $$

$\begin{aligned} \lim _{n \rightarrow \infty} \sqrt{n-1} \int_{\frac{1}{n^2}}^{\frac{1}{n}} \sqrt{\frac{1-t}{t} d t} \leq \lim _{n \rightarrow \infty} & f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq \lim _{n \rightarrow \infty} \sqrt{n}-1 \frac{2}{\sqrt{n}}\end{aligned}$

$\Rightarrow \lim\limits_{n \rightarrow \infty} \frac{\int_{\frac{1}{n^2}}^{\frac{1}{n}} \sqrt{\frac{1-t}{t}} d t}{\frac{1}{\sqrt{n-1}}} \leq \lim\limits_{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq 2$

$\begin{gathered}\Rightarrow \frac{\lim\limits_{n \rightarrow \infty} \frac{-1}{n^2} \sqrt{n-1}+\frac{2}{n^3} \sqrt{n^2-1}}{\frac{1}{2(n-1)^{\frac{3}{2}}}} \leq \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right) \leq 2\end{gathered}$

$\begin{aligned} & \therefore \lim _{n \rightarrow \infty} \frac{2(n-1)^2}{n^2}-\frac{4(n-1)^{\frac{3}{2}} \sqrt{n^2-1}}{n^3}=2 \\\\ & \therefore \lim _{n \rightarrow \infty} f\left(\frac{1}{n}\right) g\left(\frac{1}{n}\right)=2 \text { (Using Sandwich Theorem) }\end{aligned}$19:T6cb,$\begin{aligned} & f:(0,1) \rightarrow R, \\\\ & f(x)=\sqrt{n}, x \in\left[\frac{1}{n+1}, \frac{1}{n}\right), n \in N\end{aligned}$

$g:(0,1) \rightarrow R$ जहाँ

$$ \int_{x^2}^x \sqrt{\frac{1-t}{t}} d t < g(x) < 2 \sqrt{x}, x \in(0,1) $$

अब (प्रश्न के अनुसार)

$$ \begin{aligned} & \lim _{x \rightarrow \infty} f(x) \cdot g(x) \\\\ & \Rightarrow \text { रखने

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