# How do you find the limit (1/t+1/sqrtt)(sqrt(t+1)-1) as t->0^+?

Nov 28, 2016

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

#### Explanation:

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

$\therefore L = {\lim}_{t \rightarrow {0}^{+}} \left(\frac{1}{t} + \frac{1}{t} ^ \left(\frac{1}{2}\right)\right) \left({\left(t + 1\right)}^{\frac{1}{2}} - 1\right)$
$\therefore L = {\lim}_{t \rightarrow {0}^{+}} \left(\frac{1}{t} + {t}^{\frac{1}{2}} / t\right) \left({\left(t + 1\right)}^{\frac{1}{2}} - 1\right)$
$\therefore L = {\lim}_{t \rightarrow {0}^{+}} \frac{\left(1 + {t}^{\frac{1}{2}}\right) \left({\left(t + 1\right)}^{\frac{1}{2}} - 1\right)}{t}$

This is now in the indeterminate form $\frac{0}{0}$, so we can apply L'Hospital's Rule:

$\therefore L = {\lim}_{t \rightarrow {0}^{+}} \frac{\left(1 + {t}^{\frac{1}{2}}\right) \frac{d}{\mathrm{dt}} \left({\left(t + 1\right)}^{\frac{1}{2}} - 1\right) + \frac{d}{\mathrm{dt}} \left(1 + {t}^{\frac{1}{2}}\right) \left({\left(t + 1\right)}^{\frac{1}{2}} - 1\right)}{1}$
 :. L = lim_(t rarr 0^+) {( 1 + t^(1/2) ) ( 1/2(t+1)^(-1/2) ) + ( 1/2t^(-1/2) ) ( (t+1)^(1/2) - 1 ) )}
$\therefore L = {\lim}_{t \rightarrow {0}^{+}} \left\{\frac{1 + \sqrt{t}}{2 \sqrt{t + 1}} + \frac{\sqrt{t + 1} - 1}{2 \sqrt{t}}\right\}$

Let ${L}_{1} = {\lim}_{t \rightarrow {0}^{+}} \frac{1 + \sqrt{t}}{2 \sqrt{t + 1}}$, and, ${L}_{2} = {\lim}_{t \rightarrow {0}^{+}} \frac{\sqrt{t + 1} - 1}{2 \sqrt{t}}$

Then $L = {L}_{1} + {L}_{2}$

We can evaluate ${L}_{1}$ directly:

$\therefore {L}_{1} = \frac{1 + 0}{2 \sqrt{0 + 1}} = \frac{1}{2}$

And I won't do the maths because it it is really tedious, but a couple more application of L'Hospital's Rule will show that

${L}_{2} = 0$

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

We can verify this by plotting a graph of the function

$f \left(x\right) = \left(\frac{1}{x} + \frac{1}{\sqrt{x}}\right) \left(\sqrt{x + 1} - 1\right)$

graph{( 1/x + 1/sqrt(x) )( sqrt(x+1) - 1 ) [-0.811, 1.3165, -0.191, 0.873]}