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As written in the title I want to show that $f$ is continuous at the point $x_0=1$ using the $\epsilon$-$\delta$ definition.

Here's my attempt, but I am not sure it's correct.

Let $\epsilon>0$ and define $\delta=\epsilon$, then $\forall x$ sucht that |$x-x_0$|$ \leq \delta$ we have

|$f(x)-f(x_0)$|$=$|$\frac{\sqrt{2x}}{\sqrt{x+1}}-1$|$=$|$\sqrt{\frac{2x}{x+1}}-1$|$=$|$\frac{(\sqrt{\frac{2x}{x+1}}-1)(\sqrt{\frac{2x}{x+1}}+1)}{(\sqrt{\frac{2x}{x+1}}+1)}$|$=$|$\frac{\frac{x-1}{x+1}}{(\sqrt{\frac{2x}{x+1}}+1)}$|$=$$\frac{|x-1|}{|(x+1)(\sqrt{\frac{2x}{x+1}}+1)|}$

Now since the denominator is greater than one, I get

$\frac{|x-1|}{|(x+1)(\sqrt{\frac{2x}{x+1}}+1)|} \leq |x-1| \leq \delta = \epsilon$

Hence, $f$ is continuous in the point $x_0=1$.

Thank you for your help

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  • $\begingroup$ Yes, that seems okay to me; well done :) $\endgroup$ – Shaun Dec 15 '18 at 11:57
  • $\begingroup$ Yeah I can't find anything wrong here. $\endgroup$ – Anik Bhowmick Dec 15 '18 at 11:57
  • $\begingroup$ While it is clear that $x \ge 0$ because of the term $2x$ under the square root, you may want to clearly state that e.g. $\delta < \frac{1}{2}$ to safely conclude that $|x+1|\ge1$. This is nit-picking, though. $\endgroup$ – tonychow0929 Dec 15 '18 at 12:01
  • $\begingroup$ If it concerns a function $f(x)=\frac{g(x)}{h(x)}$ with $h(x_0)\neq0$ then you can prove that $g(x)$ and $h(x)$ are both continuous at $x_0$. There is a theorem that says that in that case also $f(x)$ is continuous at $x_0$. $\endgroup$ – drhab Dec 15 '18 at 12:03
  • $\begingroup$ Yes of course. This is an exercice I had to hand out and it was specified to use the definition of continuity. $\endgroup$ – Alain Dec 15 '18 at 12:09

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