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I'm trying to determine whether the following function is continuous: $y=\frac{1}{\sqrt{1+x}-\sqrt{1-x}}$. Imo it is continuous, because it is a composition of two continuous functions $y=\frac{1}{x}$ and $y=\sqrt{1+x}-\sqrt{1-x}$.

So my answer is, that the function is continuous on it's domain and it's not continuous on $\mathbb{R}$. Am I correct?

Thanks

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Exactly, in its domain of definition it is continuous (good exercise: try to find it). But of course not in all $\mathbb{R}$, since $\frac{1}{x}$ is defined only in $\mathbb{R}\setminus\{0\}$ and $\sqrt{x}$ makes sense only for $x\geq 0$.

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  • $\begingroup$ Thanks! So in order to investigate the continuity in $\mathbb{R}$, I have to find the domain of the composition and from there I can find easily points in which the function is not continuous? $\endgroup$
    – Peter F.
    Oct 20 '19 at 12:23
  • $\begingroup$ Exactly. Try first to understand when $\sqrt{1+x}-\sqrt{1-x}$ makes sense, and in order to compose it with $\frac{1}{x}$, when it it equal to zero. Then, intersect the two conditions on $x$ to obtain the domain of definition. $\endgroup$
    – Duca_Conte
    Oct 20 '19 at 12:27
  • $\begingroup$ To be more precise, rewrite the function multiplying all by $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$ in order to get rid of the square roots at the denominator. $\endgroup$
    – Duca_Conte
    Oct 20 '19 at 12:29
  • $\begingroup$ In fact if the denominator go to infinity, maybe the function still makes sense, since it can be defined equal to zero in that point. $\endgroup$
    – Duca_Conte
    Oct 20 '19 at 12:30
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You are correct but a more detailed answer would explain the domain as well. The domain is $$[-\sqrt 2 /2,0)\cup (0, \sqrt 2/2]$$

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