1.limn→∞{1n+m+1n+2m+…+1n+nm}=\displaystyle\lim_{n\to\infty}\left\{\frac1{n+m}+\frac1{n+2m}+\ldots+\frac1{n+nm}\right\}=n→∞lim{n+m1+n+2m1+…+n+nm1}=a.loge(1+m)m\frac{\log_e(1+m)}{m}mloge(1+m)b.loge(1+m)1+m\frac{\log_e(1+m)}{1+m}1+mloge(1+m)c.loge(1+m)m\frac{\log_e(1+m)}{m}mloge(1+m)d.loge(1+m)1−m\frac{\log_e(1+m)}{1-m}1−mloge(1+m)Login to continueOnly logged in users canattempt or see the solution.