# Is the following substitution legitimate?

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 $$const.b_2 = const .f_1(x) \left( b_2 + \frac{a} {f_2(x)} \right) ?$$

• 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.$f_1(x)(b_2 + \frac{a}{f_2(x)})$" – Deven Ware Sep 23 '11 at 4:16
• 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. – Ted Sep 23 '11 at 4:36
• 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? – ganzewoort Sep 23 '11 at 4:39
• Ted, but the first equation shows that $b_1$ depends on $x$, doesn't it? Therefore, Deven's reply holds. – ganzewoort Sep 23 '11 at 4:42
• @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 :) – Deven Ware Sep 23 '11 at 4:56