Evaluate limx→2f(x) (if it exists), where f(x)
=⎧⎪⎨⎪⎩x−[x],x<24,x=23x−5,x>2
limx→2−f(x)=limx→2−{x−[x]}
=limx→2−x−limx→2−[x]
=2−1=1[∵limx→k−[x]=k−1]
limx→2+f(x)=limx→2+(3x−5)
=3(2)-5
=6-5
=1
Thus,limx→2−f(x)=1=limx→2+f(x)
⇒limx→2f(x)=1
Find k so that limx→2 f(x) may exist, where f(x)={2x+3,x≤2x+k,x>2
Evaluate limx→0f(x), where f(x)={|x|x,x≠00,x=0
Let f(x)={x+5,ifx>0x−4,ifx<0 Prove that limx→0 f(x) does not exist.
Let f(x)={x+1,ifx>0x−1,ifx<0 Prove that limx→0 f(x)does not exist.