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

Question

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

Samantha Adkison · Accepted Answer

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)#

Elijah Abram · Answer

undefined To find the limit using L'Hospital's Rule, we first notice that as $ t $ approaches $ 0^+ $, both $ \frac{1}{t} $ and $ \sqrt{t} $ approach infinity. By applying L'Hospital's Rule, we differentiate the numerator and the denominator separately until we can evaluate the limit.

First, we differentiate the numerator and the denominator:
$$ \lim_{t \to 0^+} \frac{1/t + 1/\sqrt{t}}{\sqrt{t+1} - 1} $$
$$ = \lim_{t \to 0^+} \frac{-1/t^2 - 1/(2t\sqrt{t})}{1/(2\sqrt{t+1})} $$

Then, we simplify:
$$ = \lim_{t \to 0^+} \frac{-2\sqrt{t+1} - \sqrt{t}}{t^{3/2}} $$

As $ t $ approaches $ 0^+ $, both $ \sqrt{t+1} $ and $ \sqrt{t} $ approach 1. Therefore, the limit is:
$$ = -2 \times 1 = -2 $$

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

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