# Very basic question

1. Dec 15, 2013

### Husaaved

h(t) = $\sqrt{t - 1}$
h(t + Δt) = $\sqrt{t + Δt - 1}$
h(t + Δt) - h(t) = $\sqrt{t + Δt - 1}$ - $\sqrt{t - 1}$

So far so good. This is where I get confused:

$\frac{h(t + Δt) - h(t)}{Δt}$ = $\frac{1}{\sqrt{t + Δt - 1} - \sqrt{t - 1} }$

I don't understand why dividing both sides by Δt allows for this statement to be true. Can someone explain this to me? It would be very much appreciated.

Thanks a lot.

2. Dec 15, 2013

### lurflurf

That should be
$\frac{h(t + Δt) - h(t)}{Δt}$ = $\frac{1}{\sqrt{t + Δt - 1} + \sqrt{t - 1} }$
to show it observe that
$$\frac{h(t + Δt) - h(t)}{Δt} = \frac{\sqrt{t + Δt - 1} - \sqrt{t - 1} }{Δ t} \cdot \frac{\sqrt{t + Δt - 1} + \sqrt{t - 1}}{\sqrt{t + Δt - 1} + \sqrt{t - 1}} =\frac{1}{\sqrt{t + Δt - 1} + \sqrt{t - 1} }$$
That is often called rationalizing the numerator, but in this case we don't really care that the numerator is rational we want to have a sum instead of a difference.

3. Dec 15, 2013

### Husaaved

Thank you very much for your response, sorry for the typo.