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

Jan 3, 2017

I got $\frac{1}{2}$.

Try getting common denominators in the left product.

${\lim}_{t \to {0}^{+}} \left(\frac{1}{t} + \frac{1}{\sqrt{t}}\right) \left(\sqrt{t + 1} - 1\right)$

$= {\lim}_{t \to {0}^{+}} \left(\frac{1}{t} + \frac{\sqrt{t}}{t}\right) \left(\sqrt{t + 1} - 1\right)$

$= {\lim}_{t \to {0}^{+}} \frac{1}{t} \left(\sqrt{t} + 1\right) \left(\sqrt{t + 1} - 1\right)$

$= {\lim}_{t \to {0}^{+}} \frac{\sqrt{t \left(t + 1\right)} - \sqrt{t} + \sqrt{t + 1} - 1}{t}$

Now, as the numerator and denominator are both continuous in the required interval ($t > 0$), you can apply L'Hopital's rule to get:

$= {\lim}_{t \to {0}^{+}} \frac{\frac{d}{\mathrm{dt}} \left[\sqrt{{t}^{2} + t}\right] - \frac{d}{\mathrm{dt}} \left[\sqrt{t}\right] + \frac{d}{\mathrm{dt}} \left[\sqrt{t + 1}\right] - \frac{d}{\mathrm{dt}} \left[1\right]}{\frac{d}{\mathrm{dt}} \left[t\right]}$

$= {\lim}_{t \to {0}^{+}} \frac{2 t + 1}{2 \sqrt{{t}^{2} + t}} - \frac{1}{2 \sqrt{t}} + \frac{1}{2 \sqrt{t + 1}}$

$= {\lim}_{t \to {0}^{+}} \frac{2 t + 1}{2 \sqrt{{t}^{2} + t}} - \frac{\sqrt{t + 1}}{2 \sqrt{{t}^{2} + t}} + \frac{1}{2 \sqrt{t + 1}}$

This still has two terms that either go to $\infty$ or $- \infty$, so we must press on.

$= {\lim}_{t \to {0}^{+}} \frac{\frac{d}{\mathrm{dt}} \left[\left(2 t + 1 - \sqrt{t + 1}\right)\right]}{2 \frac{d}{\mathrm{dt}} \left[\sqrt{{t}^{2} + t}\right]} + \frac{1}{2 \sqrt{t + 1}}$

$= {\lim}_{t \to {0}^{+}} \frac{2 - \frac{1}{\sqrt{t + 1}}}{2 \cdot \frac{2 t + 1}{2 \sqrt{{t}^{2} + t}}} + \frac{1}{2 \sqrt{t + 1}}$

$= {\lim}_{t \to {0}^{+}} \frac{\left(2 - \frac{1}{\sqrt{t + 1}}\right) \left(2 \sqrt{t \left(t + 1\right)}\right)}{2 \cdot \left(2 t + 1\right)} + \frac{1}{2 \sqrt{t + 1}}$

$= {\lim}_{t \to {0}^{+}} \frac{2 \sqrt{t \left(t + 1\right)} - \sqrt{t}}{2 t + 1} + \frac{1}{2 \sqrt{t + 1}}$

Finally we have all denominators not going to $0$; we can plug in $0$ to get:

$\implies {\cancel{\frac{2 \left(0\right) - \left(0\right)}{2 \left(0\right) + 1}}}^{0} + \frac{1}{2 \sqrt{\left(0\right) + 1}}$

$= \textcolor{b l u e}{\frac{1}{2}}$