Why is my $\epsilon$-$\delta$ proof incorrect?

Question

Prove that $f(x)= \sqrt{4+x^2}$ is continuous at $x_0$ using the $\epsilon -\delta$ definition of continuity.

Textbook proof:

$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\underbrace{\frac{|x+x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}}_{K(x,x_0)}|x-x_0|$

$K(x,x_0)=\frac{|x+x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}\leq \frac{|x|+|x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}+\frac{|x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}$

$\leq\frac{|x|}{\sqrt{4+x^2}}+\frac{|x_0|}{\sqrt{4+{x_0}^2}}$

$\frac{|x|}{\sqrt{4+x^2}}\leq 1$ and $\frac{|x_0|}{\sqrt{4+{x_0}^2}}\leq 1\Rightarrow K(x,x_0)\leq 2\Rightarrow |f(x)-f(x_0)|\leq2|x-x_0|<2\delta=\epsilon$

So, for all $\epsilon$ we can choose a $\delta$ such that $\delta=\frac{\epsilon}{2}$.

My proof:

$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}\leq |x^2-{x_0}^2|$

$=|x-x_0||x-x_0+2x_0|<\delta^2+2|x_0|\delta$

Case $\delta<1$:

$\delta^2+2|x_0|\delta<\delta(1+2|x_0|)=\epsilon$

$\Rightarrow$ If $1+2|x_0|>\epsilon$ we have $\delta=\frac{\epsilon}{1+2|x_0|}$

Case $\delta=1$:

$\Rightarrow$ If $1+2|x_0|\leq\epsilon$ we have $\delta = 1$

Why do I get a different result for delta? What did I do wrong?

Answer 1

Regarding your proof, there is nothing wrong with it. Only that dealing with $\delta=1$ is not needed, as you can always request $\delta<1$.

As for why the "results" are different? There is no result. No one says that for a given $f$ and a given $x_0$ and a given $\varepsilon$, there is a single $\delta$. Any $\delta$ less than a given $\delta$ will work, so there is not obvious choice of a "best" $\delta$ or anything like that.

Answer 2

As others have pointed out the things that matter, I provide an argument for your reference; it is more succinct:

Let $x_{0} \in \Bbb{R}$. If $x \in \Bbb{R}$, then $$ |\sqrt{4+x^{2}} - \sqrt{4+x_{0}^{2}}| = \frac{|x-x_{0}||x+x_{0}|}{\sqrt{4+x^{2}} + \sqrt{4 + x_{0}^{2}}} < |x-x_{0}||x+x_{0}|; $$ if in addition $|x-x_{0}| < 1$, then $|x| - |x_{0}| \leq |x-x_{0}| < 1$, implying $|x+x_{0}| \leq |x|+|x_{0}| < 1 + 2|x_{0}| =: M_{0}$, implying $|x-x_{0}||x+x_{0}| < |x-x_{0}|M_{0}$; given any $\varepsilon > 0$, we have $|x - x_{0}|M_{0} < \varepsilon$ if in addition $|x-x_{0}| < \varepsilon/M_{0}$. All in all, for every $x_{0} \in \Bbb{R}$ and every $\varepsilon > 0$, it holds that $|x-x_{0}| < \min \{ 1, \varepsilon/M_{0} \}$ implies $|f(x) - f(x_{0})| < \varepsilon$.