Question

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

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

Answer 1

Answer: (A)

L.H.L. $=\underset{x\to -1^{-}}{\lim }\frac{1}{\sqrt{|x|-\{-x\}}}$
$=\underset{x\to -1^{-}}{\lim }\frac{1}{\sqrt{-x-(x+2)}}$
$=\underset{x\to -1^{-}}{\lim }\frac{1}{\sqrt{-2x-2}}=\infty$
R.H.L. $=\underset{x\to -1^{+}}{\lim }\frac{1}{\sqrt{|x|-\{-x\}}}$
$=\underset{x\to -1^{-}}{\lim }\frac{1}{\sqrt{-x-(x+1)}}$
$=\underset{x\to -1^{-}}{\lim }\frac{1}{\sqrt{-2x-1}}=1$
Hence, the limit does not exist.