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.


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

  • $\begingroup$ 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

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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