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Suppose that $$ b_1 = const .f_1(x) \left( b_2 + \frac{a} {f_2(x)} \right) $$ becomes $$ b_1 = const. b_2 $$ because $f_1(x) \rightarrow 1$ and $f_2(x) \rightarrow \infty$ when $x \rightarrow \infty$. Do you agree that it allows us to write $$ const.b_2 = const .f_1(x) \left( b_2 + \frac{a} {f_2(x)} \right) ?$$

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    no, because what if $x$ was $5$ (or anything else) then $f_2(x)$ wouldn't be $\infty$ and $f_1(x)$ wouldn't be $1$ and the equality wouldn't hold, but you could write "const.$b_2 = \lim_{x\to\infty}$ const.$f_1(x)(b_2 + \frac{a}{f_2(x)})$"2011-09-23
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    If $b_1$, $b_2$, $a$ are constants (independent of $x$), and $f_1$ and $f_2$ are continuous, and the given equation is supposed to hold for *all* $x$, then it's okay.2011-09-23
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    Thanks a bunch, Deven. This appeared to be something akin to the boundary value problems in differential equations. How would you comment the seeming similarity of the above incorrect substitution to the boundary value problems?2011-09-23
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    Ted, but the first equation shows that $b_1$ depends on $x$, doesn't it? Therefore, Deven's reply holds.2011-09-23
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    @ganzewoort I'm sorry but I don't know exactly what a "boundary value problem" is since I've never studied differential equations but hopefully someone else can help you on that :)2011-09-23

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