<s>[INST] Show that $g(x) = \sqrt{1 + x^2}$ is continuous 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?
 [/INST] 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 $|x-p|$.
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