Show that $g(x) = \sqrt{1 + x^2}$ is continuous

Question

Show that $g(x) = \sqrt{1 + x^2}$ is continuous on $\mathbb{R}$ using the $\epsilon - \delta$ argument.

This is what I was thinking...

Let $p \in \mathbb{R}$. Then $g(x)-g(p) = \sqrt{1 + x^2} - \sqrt{1 + p^2}$

$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{x^2-p^2}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} = \frac{(x+p)(x-P)}{\sqrt{1 + x^2} + \sqrt{1 + p^2}}$.

Thus, $g(x) - g(p)| = \frac{|x+p||x-p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}}$

$ \quad \quad \quad \quad \quad \quad \leq \frac{|x| + |p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} |x -p|$ $= (\frac{|x|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} + \frac{|p|}{\sqrt{1 + x^2} + \sqrt{1 + p^2}} ) |x-p| $

$ \quad \quad \quad \quad \quad \quad \leq (\frac{|x|}{\sqrt{1 + x^2}} + \frac{|p|}{\sqrt{1 + p^2}} ) |x-p|$

$ \quad \quad \quad \quad \quad \quad \leq (1+1) |x-p| = 2|x-p|$

Choose $\epsilon > 0$. Then $|g(x)-g(p)| < \epsilon$ whenever $|x-p| < \frac{\epsilon}{2}$. Let $\delta = \frac{\epsilon}{2}> 0$. Then for $\epsilon > 0$ there exists $\delta > 0$ such that $|g(x)-g(p)| < \epsilon$ whenever $|x-p|< \delta$. Hence, $g$ is continuous at $p \in \mathbb{R}$. Since $p$ is an arbitrary real number, $g$ is continuous on $\mathbb{R}$.

Is this correct?

Answer 1

A slightly "different" approach is to use ( to abuse to be correct ) the big brother inequality called Minkowski inequality of the form: $\sqrt{(a+b)^2 + (c+d)^2} \le \sqrt{a^2+c^2} +\sqrt{b^2+d^2}$. Use this inequality with $a = b = 1, c = x, d = p$ as in your answer, the follows it with the popular trigonometric substitution $\dfrac{|x|}{\sqrt{1+x^2}} = \sin \theta\le 1$, and the other part is the same which is $|x-p|$.

Answer 2

You can also follow this slightly shorter argument:

Fix an epsilon $\epsilon$.

$$|g(x)-g(p)|=|\sqrt{1+x^2}-\sqrt{1+p^2}|\leq \sqrt{|1+x^2-1-p^2|}\leq \sqrt{|x^2-p^2|}=\sqrt{|x+p||x-p|}$$

You can bound $|x+p|$ by making sure $|x-p|<1$ or $|x+p|<1+2p$. Also, you make sure that $|x-p|<\frac{\epsilon ^2}{1+2p}$ So that you have when $|x-p|<\min(1,\frac{\epsilon ^2}{1+2p})$ $$\sqrt{|x+p||x-p|}<\epsilon$$