= ltx->0 x sinx
Let Y = Xsinx
log Y = sinx log x ; // apply log on both the sides
ltx -> 0 logY = ltx->0 sinx logx ; // apply limit on both sides
logY = ltx-> 0 logx/cosecx; // infinity / infinity ; so apply L Hospital rule ; ltx -> 0 logY = log Y why because here log Y is constant doesn't depend upon x.
logY = ltx->0 (1/x)/-cosecx.cotx ; //
logY = ltx->0(-sinx. tanx)/x ; //
logY = ltx->0 -sinx. (tanx/x) ;
logY = ltx->0 -sinx . 1 ; // ltx->0 tanx/x = 1
log Y = 0
Y = e0 = 1
therefore, Y = 1