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

1 Answer
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)