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Question

I am working out the initial value for my function of s.

In the second line of working, every term in the numerator is divided by s^3, but every term in the denominator is divided by s (and not s^3).

Surely this changes the value of the fraction, as algebra normally doesn't allow you to do something like this.

Why does this work?

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    Is this allowed due to every s term being evaluated at infinity? Hence s^2 is practically s^3? – 2017-01-04
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    Each of the three brackets in the denominator is divided by $s$, so the whole denominator has been divided by $s^3$. – 2017-01-04

1 Answers 1

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Each of the three brackets in the denominator is divided by $s$, so the whole denominator has been divided by $s^3$: $$ \frac{(3s+1)(s+3)(s+100)}{s^3} = \frac{(3s+1)}{s}\frac{(s+3)}{s}\frac{(s+100)}{s} = (3+s^{-1})(1+3s^{-1})(1+100s^{-1}). $$