Question

$\underset{x\stackrel{-}{\to }1}{lim}\frac{1}{\sqrt{|x|-\{-x\}}}$ (where { x } denotes the fractional part of x) is equal to

• A. does not exists
• B. $1$
• C. $\infty$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

does not exists

Given $\underset{x\to 1}{lt}\frac{1}{|x|-\{-x\}}$

We know that

$\left\{x\right\}=x-\left[x\right]$ where $\left[x\right]$ greatest integers of $x$

$\Rightarrow \{-x\}=-x-\left[x\right]$

$\Rightarrow \{-x\}=x+[-x]$

$=x-1-\left[x\right]$

$\underset{x\to 1}{lt}\frac{1}{\sqrt{|x|+x-1-\left[x\right]}}$

for $x\to 1|x|=x$

$\Rightarrow \underset{x\to 1}{lt}\frac{1}{\sqrt{x+x-1-\left[x\right]}}$

$\Rightarrow \underset{x\to 1}{lt}\frac{1}{\sqrt{2x-1-\left[x\right]}}$

L.H.L

$\underset{x\to 1}{lt}\frac{1}{\sqrt{2x-1-1}}=\underset{x\to 1}{lt}\frac{1}{\sqrt{2x-2}}=\frac{1}{\sqrt{2\left(1\right)-2}}=\infty [\because x\to 1^{+}\Rightarrow x>1\Rightarrow [x]=1]$

R.H.L

$\underset{x\to 1^{-}}{lt}\frac{1}{\sqrt{2x-1-0}}=\frac{1}{\sqrt{2\left(1\right)-1}}=1[\because x\to 1^{-}\Rightarrow x<1\Rightarrow [x]=0]$

$\because L.H.L\ne R.H.L$

$\underset{x\to 1}{lim}\frac{1}{|x|-\{-x\}}$