Question

$\underset{x\to 3}{lim}\left(\right[x-3]+[3-x]-x),\text{where[.]denote the greatest function, is equal to:}$

• A. 4
• B. -4
• C. 0
• D. Does not exist

Answer 1

The correct option is B

-4

$R.H.L=\underset{h\to 0}{lim}\left(\right[(3+h)-3]+[3-(3+h)-(3+h))=\underset{h\to 0}{lim}\left(\right[h]+[-h]-3-h)=\underset{h\to 0}{lim}(0-1-3-h)=-4R.H.L=\underset{h\to 0}{lim}\left(\right[(3-h)-3]+[3-(3-h)-(3-h))=\underset{h\to 0}{lim}\left(\right[-h]+[h]-3+h)=\underset{h\to 0}{lim}(-1+0-3+h)=-4\text{L.H.L=R.H.L}\therefore \underset{x\to 3}{lim}\left(\right[x-3]+[3-x]-x)=-4$