I am having trouble showing that a function is continuous using the $\epsilon, \delta$ definition.
My function $f:\mathbb{R}\rightarrow(-1,1)$ is defined as $f(x) = \frac{x}{1+|x|}$.
Looking at the graph, the function is clearly continuous, but when I expand I don't get anywhere:
$ |f(x) – f(x_0)| = |\frac{x}{1+|x|} – \frac{x_0}{1+|x_0|}|
= |\frac{x(1+|x_0|) – x_0(1+|x|) }{(1+|x|)(1+|x_0|)}|$
This does not seem to simplify much at all, I cannot find a $\delta$ that satisfies:
$\forall \epsilon > 0, \exists \delta > 0, \forall x,y, \in \mathbb{R}, |x-x_0| < \delta \Rightarrow |f(x) – f(x_0)| < \epsilon $
Am I missing something here? Any advice is greatly appreciated.
Best Answer
For what follows I will need this inequality:
$$||x| - |y|| \leq |x - y|$$
and now let $\delta$ be fixed. We will see that $|x - x_0| < \delta$ will make $|f(x) - f(x_0)|$ quite small and controllable, and then we will reverse-engineer the $\delta$ we would need for an arbitrary $\epsilon$:
Start just like you did, but then do this in the numerator:
$$x(1 + |x_0|) - x_0(1 + |x|) = x(1+|x_0|) - x_0(1 + |x_0|) + x_0(1 + |x_0|) - x_0(1 + |x|)$$
Now we rewrite the numerator:
$$x(1+|x_0|) - x_0(1 + |x_0|) + x_0(1 + |x_0|) - x_0(1 + |x|) = (x- x_0)(1 + |x_0|) + x_0(|x_0| - |x|)$$
and plug it in the fraction you had:
$$\frac{|(x- x_0)(1 + |x_0|) + x_0(|x_0| - |x|)|}{|(1 + |x|)(1 + |x_0|)|}$$
use the triangular inequality to split the numerator into
$$\frac{|(x- x_0)(1 + |x_0|) + x_0(|x_0| - |x|)|}{|(1 + |x|)(1 + |x_0|)|} \leq \frac{|(x- x_0)(1 + |x_0|)|}{|(1 + |x|)(1 + |x_0|)|} + \frac{|x_0(|x_0| - |x|)|}{|(1 + |x|)(1 + |x_0|)|}$$
Now, on the one hand,
$$\frac{|(x- x_0)(1 + |x_0|)|}{|(1 + |x|)(1 + |x_0|)|} = \frac{|(x- x_0)|}{|(1 + |x|)|} \leq \delta$$
and on the other hand,
$$ \frac{|x_0(|x_0| - |x|)|}{|(1 + |x|)(1 + |x_0|)|} \leq \frac{|(|x_0| - |x|)|}{|(1 + |x|)|} \leq \frac{|x_0 - x|}{|(1 + |x|)|} \leq \delta$$
Thus $|x - x_0| < \delta \implies |f(x) - f(x_0)| < 2\delta$.
This means that, given an $\epsilon$, just take $\delta = \epsilon/2$.