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By definition:

Suppose $lim x_n = x_0$ $lim f(x_n)$ = $lim \sqrt{2x_n+1}$ = $\sqrt{2[lim (x_n)] +1}$ = $\sqrt{2x_0 +1}$ = $f(x_0)$

By epsilon-delta property:

Let $\epsilon > 0$. We want $|f(x) - f(x_0)| < \epsilon$, while $|x-x_0|<\delta$.

$|f(x)-f(x_0)|$ = $|\sqrt{2x+1}-\sqrt{2x_0+1}|$ = $|\sqrt{2x+1}-\sqrt{2(-0.5)+1}$=$\sqrt{2x+1}<\epsilon$

I think this is a good start (but correct me if it's not), but I am clueless as to what to do from here.

Just looking for tips and corrections if need be. Please do not solve for me.

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Well you’re trying to prove continuity, but at the same time you use continuity of square root when you write $lim \sqrt{2x_n+1}=\sqrt{2lim x_n +1}$

I don’t think it’s good

But the second proof when you use epsilon-delta seems to be more correct

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  • $\begingroup$ All you left to do is show that $\sqrt{2x+1}<\varepsilon$ when $x$ is close to $x_0=0.5$, but I think it’s quite clear $\endgroup$ – Anton Zagrivin Nov 20 '18 at 21:06
  • $\begingroup$ So would I set my $\delta$ to be $\frac{\epsilon^2 -1}{2}$ ? $\endgroup$ – Elizabeth Austin Griffith Nov 20 '18 at 21:15
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For $\epsilon-\delta$ proofs it is often quite useful so set

  • $h = x-x_0$. In your case this is $h = x+\frac{1}{2}$.
  • Note that $|x-x_0| < \delta \Leftrightarrow |h|<\delta$.

Now, check what happens to $|g(x) - g(x_0)|$ while trying to isolate a useful expression in $h$.

In your case there is an additional contraint as the expression under the root should not be negative. So, you have $x \geq -\frac{1}{2} \Leftrightarrow h \geq 0$.

Now, you get $$|\sqrt{2x+1} - \sqrt{2x_0+1}| \stackrel{x = h-\frac{1}{2}, \sqrt{2x_0+1} = 0}{=} \sqrt{2\left(h-\frac{1}{2} \right)+1} = \sqrt{2h} \stackrel{!}{<} \epsilon \Rightarrow h < \frac{\epsilon^2}{2}$$

It follows immediately that $\boxed{\delta = \frac{\epsilon^2}{2}}$ does it because $$\sqrt{2h}< \sqrt{2\delta} = \sqrt{2\frac{\epsilon^2}{2}}= \epsilon$$ $$$$

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