Let f(x)={x2k(x2−4)2−xwhen x is an int eger otherwise then limx→2f(x)
exists only when k=1
exists for every real k
exists for every real k = 1
does not exist
limx→2+f(x)=limx→2+k(x2−4)2−x =limx→2+k(−2−x)=−4klimx→2+f(x)=limx→2+k(x2−4)2−x=limx→2−k(−2−x)=−4k
Hence limits exists for every real value of k.
Let f(x) = 4 and f’(x) = 4. Then limx→0xf(2)−2f(x)x−2 is given by