We need to substitute 2+h for x in the limit function, where h is a tiny discrepancy in the x value.
4x-3 becomes f(2+h)=4(2+h)-3=8+4h-3=5+4h. We are asked if the discrepancy in the result 5 is 1 unit, then what is the corresponding discrepancy in x, so 4h=1 and h=0.25 or ¼. So x can have values 2±0.25, which means as low as 2-0.25=1.75 and as high as 2+0.25=2.25. Answer c describes the situation: x must be within 0.25 units of 2.