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$$ \left( a,b \in \mathbb R_{> 0} \right) ~~\wedge~~ \left( a < b\right) ~~\wedge~~ \left( \left( b-a \right) \ll a \right) $$

I want to derive the below appoximation equation .

$$ \frac{ a+b }{ b-a } =\frac{ 2a+b-a }{ b-a } \approx \frac{ 2a }{ b-a } $$

What I tried to derive it are as below .

$$ \frac{ a+b }{ b-a } =\frac{ 2a+b-a }{ b-a } $$

$$ = \frac{ 2 a }{ b-a } + \frac{ b-a }{ b-a } $$

$$ = 2 \underbrace{\left( \frac{ a }{ b-a } \right)}_\text{very large} +1 $$

$$ \approx \frac{ 2a }{ b-a } ~~ \leftarrow~~ \text{Previous 1 in the previous eqn is trifiling so it was removed} $$

Is this derivation right?

electrical apprentice
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1 Answers1

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It always depends on what you mean by '$\approx$'. Your expression $$\frac{ a+b }{ b-a } \approx \frac{ 2a }{ b-a }$$ could be interpreted as what is called asymptotically equivalent. A function $f$ is called asymptotically equivalent to another function $g$ if $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=1.$$ This would mean in your case $$\frac{ \frac{ a+b }{ b-a } }{ \frac{ 2a }{ b-a } }=\frac{ a+b }{ 2a }=\frac{1 }{2}\left(1+\frac{b}{a}\right),$$ that $b/a$ should be approximately $1$ or in other words that $a$ and $b$ should be approximately the same.

mag
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