The value of limn→∞ 1na+1na+1+1na+2+⋯+1nb is
log(ab)
log(a/b)
log(b/a)
-log(ab)
the given limit
L=limn→∞ 1na+1na+1+1na+2+⋯+1na+n(b−a)
=limn→∞ ∑r=0(b−a)n 1na+r=limn→∞ 1n∑r=0(b−a)n 1a+r/n=∫0(b−a) dxa+x=[log(a+x)]0b−a=logb−loga=log(b/a)