Question

$\underset{x\to c}{\lim }f(x)$ doesn't exist
where $[x]$, denotes step up function and $\{x\}$ fractional part function.

• A. $f(x)=[[x]]-[2x-1],c=3$
• B. $f(x)=[x]-x,c=1$
• C. $f(x)=(x{)}^2-(-x{)}^2,c=0$
• D. $f(x)=\frac{\tan (\text{sgn}x)}{\text{sgn}x},c=0$

Answer 1

Answer: (A), (B), (C), (D)

$\underset{x\to 3}{\lim }[[x]]-[2x-1]$
RHL $\underset{h\to 0}{\lim }[(3+h)]-[6+2h-1]$
As $3<3+h<4$
$\Rightarrow [3+h]=3$ and $5+2h<6$
$\Rightarrow [5+2h]=5$
$\Rightarrow \underset{h\to 0}{\lim }(3-5)=-2$
LHL $\underset{h\to 0}{\lim }[[3-h]]-[6-2h-1]$
As $2<3-h<3$
$\Rightarrow [3-h]=2$ and $4<5-2h<5$
$\Rightarrow [5-2h]=4$
$\Rightarrow \underset{h\to 0}{\lim }(2-4)=-2$
$\therefore \underset{h\to c}{\lim }f(x)$ exists, $c=3$
(B) $\underset{x\to 1}{\lim }[x]-x\underset{x\to 1^{+}}{\lim }f(x)$
$=\underset{h\to 0}{\lim }[1+h]-(1+h)$
$=\underset{h\to 0}{\lim }(1-1-h)=0$
$\underset{x\to 1^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }[1-h]-(1-h)$
$=\underset{h\to 0}{\lim }0-(1-h)=-1$
$\therefore \underset{x\to 1}{\lim }f(x)$ doesn't exists.
(C)$\underset{x\to 0}{\lim }\{x{\}}^2-\{-x{\}}^2$
$RHL\underset{h\to 0}{\lim }\{h{\}}^2-\{-h{\}}^2$
$\Rightarrow \underset{h\to 0}{\lim }(h-[h]{)}^2-((-h)-[-h]{)}^2$
$\Rightarrow \underset{h\to 0}{\lim }{h}^2-(-h+1{)}^2$
$\Rightarrow -1$
$LHL\underset{h\to 0}{\lim }\{-h{\}}^2-\{h{\}}^2$
$\Rightarrow \underset{h\to 0}{\lim }((-h)-[-h]{)}^2-(h-[H]{)}^2$
$\Rightarrow \underset{h\to 0}{\lim }(-h+1{)}^2-(h-0{)}^2$
$\Rightarrow 1$
$\therefore \underset{x\to c}{\lim }f(x);c=0$, doesnt' exist.
(d) $\underset{x\to 0}{\lim }\frac{\tan (\text{sgn}x)}{(\text{sgn}x)}$
$\Rightarrow \underset{x\to 0^{+}}{\lim },f(x)=\underset{h\to 0}{\lim }\frac{\tan 1}{1}=\tan 1$
$\underset{x\to 0^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }\frac{\tan (-1)}{-1}=\tan 1$
$\therefore \underset{x\to 0}{\lim }f(x)$ exists.