Let $h\colon (1, \infty)\to \Bbb R$ be a function. Let $h(x)= \sqrt{x^2-1}$.

Let $\epsilon>0$ be arbitrary.

Let $x_0\geq 1$ be arbitrary.

Suppose $x_0 > 1$.

Let $$\delta=\min\left\{1, \frac{\epsilon\sqrt{x_0^2-1}}{2|x_0|+1}\right\}$$

Let $x\geq1$ and $|x-x₀|<\delta$.

Thus $|h(x)-h(x_0)|=\left|\sqrt{\vphantom{x_0^2}x^2-1}-\sqrt{x_0^2-1}\right|={?}$ That's where I get stuck.

Suppose $x_0=1$.

I don't know what $\delta$ should equal.

Let $x\geq1$ and $|x-x_0|<\delta$.

Thus $|h(x)-h(1)|=\left|\sqrt{x^2-1}-\sqrt 0\right|=\left|\sqrt{x^2-1}\right|=\sqrt{|x^2-1|}={?}$ And that's where I get stuck again.

It would end with: Since $\epsilon$ is arbitrary, $h$ is continuous at $x₀$. And since $x_0$ is arbitrary, $h$ is continuous for all $x\in (1, \infty)$


rationalize by $|\sqrt{x^2-1}+\sqrt{x_0^2-1}|$


$$ |\sqrt{x^2-1}-\sqrt{x_0^2-1}| = |\sqrt{x^2-1}-\sqrt{x_0^2-1}|\frac{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|}{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|}$$

$$ = \frac{|(x^2-1)-(x_0^2-1)|}{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|} $$ $$\leq \frac{|x^2 - x_0^2|}{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|}$$ $$\leq |x-x_0|\frac{|x|+|x_0|}{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|}$$ $$\leq |x-x_0|$$ Where the last inequality is becuase on $(1,\infty)$ the denominator is always positive. So given $\epsilon>0$, let $\delta = \epsilon$, so that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$

  • $\begingroup$ This only solves when $x_0>0$ though, correct? Thank you by the way. I was getting something like that, but the end is so much simpler. So, instead of $\delta$ being that minimum thing, it would just be let $\delta=\epsilon$? $\endgroup$ – Maddy Nov 5 '13 at 8:04
  • $\begingroup$ Yes, exactly. $\delta = \epsilon$. This argument actually holds for $I = (-\infty,1)\cup(1,\infty)$. As long as the square root is not zero or imaginary. It suffices to show it on $(1,\infty)$ since the argument is $x^2$. $\endgroup$ – RDizzl3 Nov 5 '13 at 8:11
  • $\begingroup$ What about with $x_0 =0$? It seems like that would be the simpler case? $\endgroup$ – Maddy Nov 5 '13 at 8:15
  • $\begingroup$ The function actually isn't continuous at $x_0$ since $\sqrt{0-1} = \sqrt{-1} = i$ which isn't on the real line $\mathbb{R}$. So to show continuity for all $x$, $x_0$ has to be a random point on $(1,\infty)$. Proving continuity at one point does not mean that it is continuous on the interval. $\endgroup$ – RDizzl3 Nov 5 '13 at 8:19
  • $\begingroup$ You just made my day so much better. I don't know why I got caught up on trying to prove at $x_0=0$ when I didn't even include it in the interval in the first place. Thank you! :) $\endgroup$ – Maddy Nov 5 '13 at 8:27

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