How do you find #lim (1/t+1/sqrtt)(sqrt(t+1)-1)# as #t->0^+# using l'Hospital's Rule?
1 Answer
I got
Try getting common denominators in the left product.
#lim_(t->0^(+)) (1/t + 1/sqrtt)(sqrt(t + 1) - 1)#
#= lim_(t->0^(+)) (1/t + sqrtt/t)(sqrt(t + 1) - 1)#
#= lim_(t->0^(+)) 1/t(sqrtt + 1)(sqrt(t + 1) - 1)#
#= lim_(t->0^(+)) (sqrt(t(t+1)) - sqrtt + sqrt(t+1) - 1)/t#
Now, as the numerator and denominator are both continuous in the required interval (
#= lim_(t->0^(+)) (d/(dt)[sqrt(t^2 + t)] - d/(dt)[sqrtt] + d/(dt)[sqrt(t+1)] - d/(dt)[1])/(d/(dt)[t])#
#= lim_(t->0^(+)) (2t + 1)/(2sqrt(t^2 + t)) - 1/(2sqrtt) + 1/(2sqrt(t+1))#
#= lim_(t->0^(+)) (2t + 1)/(2sqrt(t^2 + t)) - sqrt(t+1)/(2sqrt(t^2 + t)) + 1/(2sqrt(t+1))#
This still has two terms that either go to
#= lim_(t->0^(+)) (d/(dt)[(2t + 1 - sqrt(t+1))])/(2d/(dt)[sqrt(t^2 + t)]) + 1/(2sqrt(t+1))#
#= lim_(t->0^(+)) (2 - 1/(sqrt(t+1)))/(2*(2t + 1)/(2sqrt(t^2 + t))) + 1/(2sqrt(t+1))#
#= lim_(t->0^(+)) ((2 - 1/(sqrt(t+1)))(2sqrt(t(t+1))))/(2*(2t + 1)) + 1/(2sqrt(t+1))#
#= lim_(t->0^(+)) (2sqrt(t(t+1)) - sqrtt)/(2t + 1) + 1/(2sqrt(t+1))#
Finally we have all denominators not going to
#=> cancel((2(0) - (0))/(2(0) + 1))^(0) + 1/(2sqrt((0)+1))#
#= color(blue)(1/2)#