Question

$\underset{x\to c}{lim}f\left(x\right)$ does not exist when (where [x] denotes step up function & {x} fractional part function).

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

Answer 1

Answer: (B), (C)

$f\left(x\right)=\left[x\right]-x,c=1$
$f\left(x\right)=\left\{x^2\right\}-\{-x{\}}^2,c=0$

Option A: f(x) = [[x]] - [2x - 1], at c = 3
Let's check right hand limit at c = 3
x = 3 + h
$\underset{h\to 0}{lim}f(3+h)=\underset{h\to 0}{lim}\left[\right[3+h\left]\right]-\left[2\right(3+h)-1]=4-6=-2$
Let's check Left hand limit at c = 3
x = 3 - h
$\underset{h\to 0}{lim}f(3-h)=\underset{h\to 0}{lim}\left[\right[3-h\left]\right]-\left[2\right(3-h)-1]=3-5=-2$
${lim}_{x\to 3}$ f(x) does exist.
Option B: f(x) = [x] - x at c = 1
Let's check right hand limit at c = 1
x = 1 + h
$\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}[1+h]-(1+h)=1-1=0$
Let's check Left hand limit at c = 1
x = 1 - h
$\underset{h\to 0}{lim}f(1-h)=\underset{h\to 0}{lim}[1-h]-(1-h)=0-1=-1$
${lim}_{x\to 1}$ f(x) does not exist.
Option C: f(x) = {x${}^2$} - {-x}${}^2$, at c = 0
Let's check right hand limit at c = 0
x = 0 + h
$\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}(h{)}^2-{-h}^2=0-1=-1$
Let's check Left hand limit at c = 0
x = 0 - h
$\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}(-h{)}^2-h^2=0-0=0$
${lim}_{x\to 0}$ f(x) does not exist.
Option D: f(x) = $\frac{tan\left(sgnx\right)}{sgnx}$,at c = 0
Let's check right hand limit at c = 0
x = 0 + h
$\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}\frac{tan\left(sgn\right(h\left)\right))}{sgn\left(h\right)}=tan\left(1\right)$
Let's check Left hand limit at c = 0
x = 0- h
$\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}\frac{tan\left(sgn\right(-h\left)\right))}{sgn(-h)}=\frac{-tan\left(1\right)}{-1}=tan\left(1\right)$
${lim}_{x\to 0}$ f(x) does exist.
Hence, Option B and C.