Find:
(i)limx→2[x]
(ii)limx→52[x]
(iii)limx→1[x]
(i)limx→2[x]
LHL:
limx→2−[X]
Let x =2-h,
⇒h=2−x
as x→20⇒x<2 slightly
⇒2−x>0⇒h>0
⇒h→0+
limh→0+[2−h]=1
RHLlimx→2+[X]
Let x=2+h,⇒h=x−2
ax x→2+⇒x>2 slightly
⇒−2>0⇒h>0
⇒h→0+
limh→0+[2+h]
LHL≠RHL
Thus, limx→2[x] does not exist.
(ii)limx→52[x]
RHL:limx→5+2[x]
Let x=52+h,
⇒h=x−52
as x→5+2⇒x>52 slightly
⇒x−52>0
⇒h>0
⇒h→0+
limh→0+[52+h]=2
Hence, limx→5−2[x]
Let x=52−h,⇒h=52−x
as x→5−2⇒x<52 slightly
⇒52−x>0
⇒h>0
⇒h→0+
limh→0+[52−h]=2
Hence, limx→5−2[x]=2
Since, limx→5−2[x]=limx→5+2[x]
⇒limx→52[x]=2
(iii)limx→1[x]
LHL:limx→1−[x]
Let x =1-h,
⇒h=1−x
as x→1−⇒x<1 slightly
⇒1−x>0⇒h>0
⇒h→0+
limh→0+[1−h]
limx→1−[x]=0
RHL:limx→1+[x]
Let x =1+h,
⇒h=x−1
as x→1+⇒x>1 slightly
⇒x−1>0⇒h>0
⇒h→0+
limx→0+[1+4]=1
⇒limx→1[x]≠limx→1+[x]
Thus, limx→1[x] does not exist.