Mathematics - Limits Question with Solution | TestHub

Given,

$f:\left(0, 1\right)\to \mathrm{ℝ}$ be the function defined as $f\left(x\right)=\sqrt{n}$ if $x\in [\frac{1}{n+1}, \frac{1}{n})$ where $n\in \mathrm{\mathbb{N}}$

And $g:\left(0, 1\right)\to \mathrm{ℝ}$ be a function such that ${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g\left(x\right)<2\sqrt{x}$ for all $x\in \left(0, 1\right)$. 

Now we need to solve $1$ sided limit here to get some answer, as  $\underset{x\to 0^{-}}{\lim }$ doesn’t exist here (not in domain)

As $\frac{1}{n+1}\le x< \frac{1}{n}$

$\Rightarrow n+1\ge \frac{1}{x}> n$

$\Rightarrow n\ge \frac{1}{x}-1\Rightarrow \sqrt{n}\ge \sqrt{\frac{1}{x}-1}$

So, let $f\left(x\right)=\sqrt{\left(\frac{1}{x}\right)-1}$ where $\left(·\right)=$ least integer function

Now multiplying $f\left(x\right)$ in ${\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt<g\left(x\right)<2\sqrt{x}$ and taking limit we get,

$\underset{x\to 0^{+}}{\lim }{\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt·\sqrt{\left(\frac{1}{x}\right)-1}\le \underset{x\to 0^{+}}{\lim }f\left(x\right)·g\left(x\right)\le \underset{x\to 0^{+}}{\lim }\sqrt{\left(\frac{1}{x}\right)-1}\times 2\sqrt{x}$

Now $\underset{x\to 0^{+}}{\lim }\sqrt{\left(\frac{1}{x}\right)-1}\times 2\sqrt{x}=\underset{x\to 0^{+}}{\lim }2\sqrt{x}\sqrt{\left[\frac{1}{x}\right]} \left(\frac{1}{x}\notin Z\right)$

$=\underset{x\to 0^{+}}{\lim }2\sqrt{x\left(\frac{1}{x}-\left\{\frac{1}{x}\right\}\right)}=2$

$=\underset{x\to 0^{+}}{\lim }2\sqrt{1-x\left(\frac{1}{x}\right)}=2; \left(\frac{1}{x}\notin Z\right)$

So, $\underset{x\to 0^{+}}{\lim }{\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt·\sqrt{\frac{1}{x}-\left\{\frac{1}{x}\right\}}=\frac{{\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt·\sqrt{1-x\left\{\frac{1}{x}\right\}}}{\sqrt{x}}$

And $\underset{x\to 0^{+}}{\lim }\frac{{\int }_{x^2}^x\sqrt{\frac{1-t}{t}}dt}{\sqrt{x}}=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{\frac{1-x}{x}}-2x\sqrt{\frac{1-x^2}{x^2}}}{\frac{1}{2\sqrt{x}}}$

{Using L-hospital rule}

$=\underset{x\to 0^{+}}{\lim }2\sqrt{1-x}-4\sqrt{x}·\sqrt{1-x^2}=2$

Similarly for $\frac{1}{x}\in Z$ is equal to $2$.