How do you find #lim (1/t+1/sqrtt)(sqrt(t+1)-1)# as #t->0^+# using l'Hospital's Rule?

1 Answer
Truong-Son N.
Jan 3, 2017

I got #1/2#.

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 (#t > 0#), you can apply L'Hopital's rule to get:

#= 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 #oo# or #-oo#, so we must press on.

#= 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 #0#; we can plug in #0# to get:

#=> cancel((2(0) - (0))/(2(0) + 1))^(0) + 1/(2sqrt((0)+1))#

#= color(blue)(1/2)#

Related questions
See all questions in Indeterminate Forms and de L'hospital's Rule
Impact of this question
1746 views around the world
You can reuse this answer
Creative Commons License