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Ive been told that if you have a expression of the form $\frac{c}{f(x)}$ where $c$ is constant and $f(x)$ can be zero at some $x$ then if $c=0$,

$\frac{0}{f(x)}=0$ no matter what the value of $f(x)$ is.

I'm not particularly convinced, could someone perhaps explain this?

For instance, given the ode $(1-x^2)y'-xy=x(1-x^2)$, it has a solution of the form $y=-\frac{1}{3}(1-x^2)+\frac{c}{\sqrt(1-x^2)}$. I am supposed to show whether a unique solution exists for $y(0)=1$. Plugging in the initial conditions, I get $0/0$ in the right hand side, but apparently c=0 is valid , may someone elaborate?

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    Surely not, as if $f(x)=0$ then $\frac{0}{f(x)}$ must be undefined. – Eric Nov 30 '19 at 14:51
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    I don't think that's true. The expression is not defined when $f(x)=0.$ – Kenta S Nov 30 '19 at 14:51
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    Since there is no proof in your question, why did you tag it as proof-verification and proof-explanation? – José Carlos Santos Nov 30 '19 at 14:51
  • @JoséCarlosSantos I suppose it was a step of one more general proof but I agree that without any other details these tags are not suited. – user Nov 30 '19 at 14:56
  • Perhaps the tags point to a concrete context, e.g. ordinary differential equations, in which the expression is meant to be a solution, maybe known to be differentiable. In any case if the roots of $f(x)$ are isolated, then knowing the expression to be continuous would suffice to "fill in" the undefined values as also necessarily zero. – hardmath Nov 30 '19 at 14:56
  • @topologicalmagician Maybe you should add some more detail for the context in order to understand better where the claim of the lecturer came from. – user Nov 30 '19 at 14:58
  • @hardmath please elaborate on that statement, I included details. –  Nov 30 '19 at 15:04
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    @topologicalmagician: Thanks, this is one of those situations in which context is essential to making the most sense of your Question. I'll expand my Comment in an answer. – hardmath Nov 30 '19 at 15:11
  • @hardmath you are correct. I also would like to thank you for being helpful. Thank you. –  Nov 30 '19 at 15:21
  • @hardmath may you please expand on it when you have the time,please? –  Nov 30 '19 at 19:10

2 Answers2

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Indeed we have that

$$\frac{0}{f(x)}=0$$

for any $x$ such that $f(x)\neq 0$, otherwise the expression is undefined.

Refer also to the related

user
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This looks like it might be an oversimplification of a different true statement. If $x$ is an isolated zero of $f$, then it is true that $\lim_{y \rightarrow x} (\frac {0}{f(y)})=0$. In some cases this might be oversimplified to what your lecturer has said.

Edit based on ODE context

Plugging in the initial conditions should not give you $c = 0$, but instead $c=\frac {4}{3}$, which seems like a simple mistake on your part. However, if c had been zero, the solution given would really be an over simplification of the relationship between the relevant solution $y_1=-\frac {1}{3}(1-x^2)$ and the more general solution $y_2=-\frac {1}{3}(1-x^2)+\frac {c}{\sqrt{1-x^2}}$. $y_1$ solves the differential equation and you can check that easily by hand. $y_2$ solves the differential equation but isn't ever technically defined for |x|>1. However, when approaching the general problem, it's often more convenient to treat $y_1$ as the sub-case of $y_2$ with $c=0$. This isn't technically true, but it's much more convenient, and aligns with the basic intuition that $\frac {0}{anything}$ is $0$.

Robo300
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