# show that $\frac{\sqrt{2x}}{\sqrt{x+1}}$ is continuous at the point $x_0=1$

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

• Yes, that seems okay to me; well done :) – Shaun Dec 15 '18 at 11:57
• Yeah I can't find anything wrong here. – Anik Bhowmick Dec 15 '18 at 11:57
• 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. – tonychow0929 Dec 15 '18 at 12:01
• 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 $x_0$. – drhab Dec 15 '18 at 12:03
• Yes of course. This is an exercice I had to hand out and it was specified to use the definition of continuity. – Alain Dec 15 '18 at 12:09