How do you find lim sqrtx/(x-1) as x->1^+ using l'Hospital's Rule or otherwise?

${\lim}_{x \to {1}^{+}} \frac{\sqrt{x}}{x - 1} = + \infty$
We cannot use l'Hospital's Rule as the limit does not present itself in one of the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
We can however note that for $x \to {1}^{+}$ the numerator is continuous, while the denominator tends to zero by positive values so that posing $x = 1 + \delta$:
${\lim}_{x \to {1}^{+}} \frac{\sqrt{x}}{x - 1} = {\lim}_{\delta \to {0}^{+}} \frac{\sqrt{1 + \delta}}{1 + \delta - 1} = {\lim}_{\delta \to {0}^{+}} \frac{\sqrt{1 + \delta}}{\delta} = \left({\lim}_{\delta \to {0}^{+}} \sqrt{1 + \delta}\right) \cdot \left({\lim}_{\delta \to {0}^{+}} \frac{1}{\delta}\right) = + \infty$