# The limit of case 2

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The limit of case 2

We have two rather algebraically differently looking solutions, an exponential but as r1r2 the limit of the second case ought to be the first case. Before you read any further, why not try to figure it out yourself?

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Ok, back again. Let us have a look at

$y = {y_0}{\left( {1 - \frac{{\left( {{r_2} - {r_1}} \right)}}{{{V_0}}}t} \right)^{\frac{{{r_2}}}{{{r_2} - {r_1}}}}}$

Now let us do the substitution

$n = \frac{{{r_2}}}{{{r_2} - {r_1}}}$

That means that

$\frac{{\left( {{r_2} - {r_1}} \right)}}{{{V_0}}} = \frac{{{r_2}/{V_0}}}{n}$

and thus that

$y = {y_0}{\left( {1 - \frac{{{r_2}/{V_0}}}{n}t} \right)^n}$

But we know that

${e^x} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{x}{n}} \right)^n}$

and thus that

$y = {y_0}{\left( {1 - \frac{{{r_2}/{V_0}}}{n}t} \right)^n} = {y_0}{e^{ - \frac{{{r_2}}}{V}t}} = {y_0}{e^{ - \frac{r}{V}t}}$

I.e. the solution we found in case 1.

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