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I am given a function where $$f:S\subset \mathbb{R}_+\rightarrow \mathbb{R} $$

$$f(x)=\frac{\sqrt{x}-1}{x-1}$$

I need to use the following definition to prove that $f$ is continuous at $z\in S$;

$f(x)$ converges to the limit, L, as $x$ tends to $z$, for every $\epsilon>0$ , there exists $\delta>0$ such that:

$|x-z|<\delta \Rightarrow |f(x)-L|<\epsilon $

I also have the following information that I need to use to prove continuity at $z\in S$: $$|\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{z}+1}|<|\sqrt{z}-\sqrt{x}|$$

I am also given a hint that $\sqrt{z}\ge0$ which implies $\sqrt{x}+\sqrt{z}\ge \sqrt{x}-\sqrt{z}$

What I have done so far is:

$\lim_{x \to z} f(x) = \frac{\sqrt{z}-1}{z-1}$

$|x-z|<\delta \Rightarrow |\frac{\sqrt{x}-1}{x-1}-\frac{\sqrt{z}-1}{z-1}|<\epsilon $

At this point, I'm not sure how to incorporate the information into proving that $f$ is continuous at $z\in S$

Forgive me, I am not the brightest mathematician, any help or hints to point me in the right direction would be appreciated it

ThomasVDV
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  • Are you able to use the fact that a quotient of continuous functions is likewise continuous, assuming that the denominator of the quotient is non-zero? Or do you need to prove this from first principles? If the latter, break up the proof into two cases: $z= 1$ and $z \neq 1$. – Robert Shore May 23 '20 at 02:38
  • Here's a more direct hint. If $x \gt 0, x-1= (\sqrt x-1)(\sqrt x + 1)$. – Robert Shore May 23 '20 at 02:40
  • Robert Shore gives a good hint. I'm just wondering, how comfortable are you with the idea of epsilon-delta definitions in general? Have you found limits of linear polynomials? Quadratic polynomials? Simple rational functions? If you're very uncomfortable with applying the epsilon-delta definition, then you might need more than hints. – user790072 May 23 '20 at 03:18
  • Thank you for your help guys :) I think I may have solved it, please see the answer I have attached. Thanks, Robert the hint helped me understand why I needed to use the information to prove f is continuous at z. – ThomasVDV May 23 '20 at 10:27

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Using the fact that if $x>0$, then $x-1=(\sqrt{x}-1)(\sqrt{x}+1)$

I need to show to show that for a given $\epsilon>0$, there exists a $\delta$ such that:

$$|x-z|<\delta \Rightarrow|\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{z}+1}|<\epsilon$$

Since

$z\ge 0$, it follows that; $\sqrt{x}-\sqrt{z}\le\sqrt{x}-\sqrt{z}$

[By multiplying the inequality on both sides by $\sqrt{x}-\sqrt{z}$]

Note: I have left out the steps of derivation to keep it short. But by multiplying on both sides we arrive at the following:

$$(\sqrt{x}-\sqrt{z})^2<x-z$$

We know that $$|x-z|<\delta$$ Then we use the information to show that $$|\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{z}+1}|<|\sqrt{x}-\sqrt{z}|<\sqrt{\delta}$$ Then by choosing a $\delta=\epsilon^2$ , we have proven that $f$ is continuous at $z\in S$ such that:

$$|x-z|<\delta \Rightarrow|\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{z}+1}|<\sqrt{\epsilon^2}=\epsilon$$

ThomasVDV
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