3
$\begingroup$

Please bear with me. So, for example the limit for $a$ in this hyperbola:

$$\lim_{a \to 0^+} (y^2 -x^2 = a) \, ,\,\,\, \, y \ge 0 \tag{A}$$

This appears to make the equation go to: $$y^2 = x^2 \, ,\,\,\, \, y \ge 0 \tag{B}$$ $$y = |x| \tag{C}$$

So, is $(A)$ a valid mathematical definition of the absolute value? And furthermore, is $(A)$ differentiable? Because if it was, that immediately means that it's differentiable as $a$ is aribitrarily close to $0$ which is the same thing as $(C)$, but we know $(C)$ is not differentiable at $0$.

In short, I'm wondering if (and maybe the above is a bad example) we can show, via a parameter, that one function converges to another in this type of limiting process and, if so, why would it lose it's differentiability (if that even makes sense)?

Here's a Desmos link link to the action I'm talking about.

$\endgroup$

1 Answer 1

2
$\begingroup$

What you've stumbled upon is a property of something called uniform convergence. Uniform convergence is a way to talk about how a sequence of functions approaches a limiting function.

A sequence of functions $f_n(x) \xrightarrow[unif]{} f(x) $ if for each $\epsilon > 0$ and $x \in I$ there exists N such that $n > N$ implies $| f_n(x) - f(x) | < \epsilon$

What this means intuitively is that our sequence of functions converges to the limiting function at a worst case rate. So for all points in the interval, the error, $| f_n(x) - f(x) |$ can be made as small as we want.

Lets prove that your sequence converges uniformly: Let $f_n(x) = \sqrt{x^2 + \frac{1}{n}}$ and let $f(x) = |x|$. If we let $n \to \infty$, then this is identical to your question, except we only take the upper branch of the hyperbola.

Let $\epsilon > 0$ be given. Let $x \in I$, where $I$ is some bounded interval. Now, choose $N = \frac{1}{\epsilon^2}$ Then $n>N$ implies $$\left|\sqrt{x^2+\frac{1}{n}}-|x|\right|=\sqrt{x^2+\frac{1}{n}}-|x|$$ Since $x^2$ and $\frac{1}{n}$ are both positive$$\sqrt{x^2+\frac{1}{n}}-|x|\leq \sqrt{x^2}+\sqrt{\frac{1}{n}}-|x|$$ But $\sqrt{x^2}=|x|$ so we have $$\sqrt{\frac{1}{n}} \leq \sqrt{\frac{1}{N}}=\epsilon$$ So the sequence converges uniformly for ANY bounded interval.

What does all this mean? It means that you are exactly right. The upper branch of that hyperbola does approach $|x|$. Uniform convergence, then, must not conserve differentiability. It does however conserve continuity and integrability.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .