R = 1 $$-$$ qⁿ

$$\Rightarrow$$ R = 1 $$-$$ (1 $$-$$ P)ⁿ

R = 1 $$-$$ $${\left[ {1 - {1 \over T}} \right]^n}$$

Given than T = n and n $$\to$$ $$\infty$$

$$\Rightarrow$$ R = 1 $$-$$ $$\left[ {\mathop {\lim }\limits_{n \to \infty } {{\left( {1 - {1 \over n}} \right)}^n}} \right]$$

$$\Rightarrow$$ R = 1 $$-$$ $$\left[ {\mathop {\lim }\limits_{n \to \infty } {{\left\{ {1 + \left( { - {1 \over n}} \right)} \right\}}^n}} \right]$$ ....... (i)

We know that,

$$\mathop {\lim }\limits_{n \to \infty } {(1 + x)^{1/x}} = e$$,

So let's take $${{ - 1} \over n} = y$$

$$\therefore$$ $$n = {{ - 1} \over y}$$

at n = $$\infty$$; y = 0

Now, $$\mathop {\lim }\limits_{n \to \infty } {\left( {1 - {1 \over n}} \right)^n}$$

$$ = \mathop {\lim }\limits_{y \to 0} {(1 + y)^{ - 1/y}}$$

$$ = \mathop {\lim }\limits_{y \to 0} {(1 + y)^{{1 \over y}( - 1)}}$$

$$ = {\left[ {\mathop {\lim }\limits_{y \to 0} {{(1 + y)}^{{1 \over y}}}} \right]^{ - 1}}$$

$$ = {e^{ - 1}}$$

Put in eq. (i)

$$\therefore$$ $$R = 1 - {e^{ - 1}}$$

$$R = 1 - {1 \over e}$$

$$R = 0.6321$$ i.e. 63.2%

Flood Estimation and Flood Routing · Hydrology · GATE CE

Marks 1