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Let $g(x)=\sqrt{x^{2}+4}$ from $\mathbb{R}\rightarrow \mathbb{R}$. I want to show that $g$ is continuous at $x=1$.

I have to show that for any $\epsilon>0$, there exists a $\delta>0$ such that $|x-1|< \delta \implies |g(x)-g(1)|<\epsilon$.

So I do some rough work first and start with $$|g(x)-g(1)|$$ $$=|\sqrt{x^{2}+4} -\sqrt{5}|$$ $$\leq|\sqrt{x^{2}+4}| + |\sqrt{5}|$$.

I am stuck on what to do after. The plan is to choose the right $\delta$.

OGC
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  • Should there be something between your lines? Maybe equal or inequal signs? –  Sep 15 '15 at 00:18
  • @Ive added the equal signs. – OGC Sep 15 '15 at 00:21
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    How about if you multiply top and bottom by conjugate $\sqrt{x^2+4}+\sqrt{5}$? That creates an $x^2-1$ in the numerator which is factorable. That could be handy since the "delta" part contains an $x-1$. Just a funny thought. – imranfat Sep 15 '15 at 00:22
  • The last one is not true at all. Those two things are not equal. –  Sep 15 '15 at 00:22
  • @avid19 Doesn't it hold by the absolute-value properties: $|-a|=|a|$ and $|a+b|=|a|+|b|$? – OGC Sep 15 '15 at 00:26
  • That second equality is patently false. For example, let $a=-b=1$ then you are claiming that $|1-1|=|1|+|-1|$, or that $0=2$. –  Sep 15 '15 at 00:28
  • @avid19 Yes, you're right. Should have an inequality in between. – OGC Sep 15 '15 at 00:31

1 Answers1

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As suggested in a comment of imranfat, multiply by the conjugate, i.e. \begin{align*} \sqrt{x^2+4}-\sqrt{5}=\frac{(\sqrt{x^2+4}-\sqrt{5})(\sqrt{x^2+4}+\sqrt{5})}{\sqrt{x^2+4}+\sqrt{5}}=\frac{x^2-1}{\sqrt{x^2+4}+\sqrt{5}}=\frac{(x-1)(x+1)}{\sqrt{x^2+4}+\sqrt{5}}. \end{align*} Now, if $|x-1|<\delta$, then $|x+1|=|x-1+2|\leq|x-1|+2\leq\delta+2$. Since $\sqrt{x^2+4}+\sqrt{5}\geq\sqrt{5}\geq1$, it follows that \begin{align*} |\sqrt{x^2+4}-\sqrt{5}|=\frac{|x-1||x+1|}{\sqrt{x^2+4}+\sqrt{5}}\leq\delta(\delta+2). \end{align*} Can you go on from here?

sranthrop
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