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I want to prove the function $f(x)=$ \begin{cases} \frac{1-\sqrt{1+ax^2}}{x^2}, & x \ne 0\\ a, & x=0 \end{cases}

( where $a \in \Bbb{R}$ ) is continuous in its domain. To begin with, $f$ is continuous on an interval if it is continuous at every point of that interval, in this case the interval $[-\infty ,\infty ]$. Lets first look at the value of $f$ when $x \neq 0$ :

$$f(x) = \frac{1-\sqrt{1+ax^2}}{x^2}$$

Lets assume there exists functions $g$ and $h$, such that $g(x)=1-\sqrt{1+ax^2}$ and $h(x) = x^2$. Now $f$ is continuous at $\Bbb{R}$\ $0$, if $g$ and $h$ are continuous there as well. And they are, since $g$ and $h$ are continuous everywhere in their domain. Therefore $f(x)$ is continuous on the interval $\Bbb{R}$\ $0$.

In the case where $x=0$, we can say $f$ is continuous there if the limit $$\lim_{x\to 0}f(x)=f(0)=f(a)$$ Which is true by the definition of $f$. Is this enough to show $f$ is continuous in its domain?

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2 Answers 2

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You must show that both limits $$\lim_{x\to 0^{+}}\frac{1-\sqrt{1+ax^{2}}}{x^{2}} \quad \mbox{and} \quad \lim_{x\to 0^{-}}\frac{1-\sqrt{1+ax^{2}}}{x^{2}}$$ are equal to $f(0) = a$. Note, however, that this function depends only on $x^{2}$ so it is enough to consider only one limit. But using L'Hospital, it is easy to conclude that: $$\lim_{x\to 0}\frac{1-\sqrt{1+ax^{2}}}{x^{2}} = -\frac{a}{2} \neq a = f(0)$$ so the function is not continuous at $x=0$.

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  • $\begingroup$ Come on, you really need L'Hospital here?! $\endgroup$
    – nonuser
    Nov 21, 2019 at 16:15
  • $\begingroup$ hahaha that was me first idea! $\endgroup$ Nov 21, 2019 at 16:24
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$$\lim_{x\to 0}\frac{1-\sqrt{1+ax^2}}{x^2}= \lim_{x\to 0}\frac{-ax^2}{x^2(1+\sqrt{1+ax^2})} = -{a\over 2}$$

So your function is not continous at $x=0$ if $a\ne 0$.

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