# lim_(t->oo)(sqrt(t)-1)/(t-1) =  ?

Sep 20, 2016

I think it was for $t \to \infty$...giving zero in the limit.

#### Explanation:

If it is $t \to \infty$:

${\lim}_{t \to \infty} \frac{\sqrt{t} - 1}{t - 1} = \frac{\infty}{\infty}$

To solve this we can use de L'Hospital Rule deriving top and bottom to get:
${\lim}_{t \to \infty} \frac{\frac{1}{2 \sqrt{t}}}{1} = {\lim}_{t \to \infty} \frac{1}{2 \sqrt{t}}$

as $t \to \infty$ the argument tends to zero;

${\lim}_{t \to \infty} \frac{1}{2 \sqrt{t}} = 0$

Sep 23, 2016

${\lim}_{x \to \infty} \frac{\sqrt{t} - 1}{t - 1} = 0$

#### Explanation:

An alternative method, solving algebraically:

${\lim}_{x \to \infty} \frac{\sqrt{t} - 1}{t - 1}$

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

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

$= {\lim}_{x \to \infty} \frac{1}{\sqrt{t} + 1}$

$= \frac{1}{\infty}$

$= 0$

Sep 23, 2016

Making $y = \sqrt{t}$

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