Q: Epsilon-delta proof for $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$

Question

In proving $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$, I have done this working:

Let $ϵ>0$.

$0<|x-1|< \Rightarrow|\frac{x}{(x^2+1)}-\frac{1}{2}|<$

$ |\frac{x}{(x^2+1)}-\frac{1}{2}|< = |\frac{2x-x^2+1}{2(x^2+1)}| = |\frac{-2(x-1)^2}{x(x-1)^2+4x}| = |\frac{-(x-1)^2}{(x-1)^2+2(x-1)+2}| = \frac{(x-1)^2}{|(x-1)^2+2(x-1)+2|}|$

after which I get stuck at trying to implement triangle inequality to push in the modulo in the denominator to get $|x-1|$ terms to substitute .

Answer 1

You have\begin{align}\frac x{x^2+1}-\frac12&=\frac{-x^2+2x-1}{2x^2+2}\\&=\frac{-(x-1)^2}{2x^2+2}.\end{align}Now, you alwas have $2x^2+2\geqslant2>1$. So$$\left|\frac{-(x-1)^2}{2x^2+2}\right|<(x-1)^2\tag1$$and therefore, given $\varepsilon>0$, if you take $|x-1|<\sqrt{\varepsilon}$, it will follow from $(1)$ that$$\left|\frac x{x^2+1}-\frac12\right|<\varepsilon.$$

Answer 2

As $ x $ goes to $ 1$, we can always assume that $$|x-1|<2$$ and we look for $ \delta$ such that $$|x-1|<2 \text{ and } |x-1|<\delta \implies$$ $$ |\frac{-(x-1)^2}{2x^2+2}|<\epsilon$$

but $$2x^2+2\ge 2$$ and $$|x-1|<2\implies $$ $$|\frac{(x-1)(x-1)}{2x^2+2}|\le |x-1|$$

So, we just need a $ \delta<2$ such that

$$|x-1|<\delta\implies |x-1|<\epsilon$$

from here, we see that we can take $$\delta=\min(2,\epsilon)$$

Answer 3

Alternative approach:

Let

$$0 < |x - 1| < \delta \leq \frac{1}{10}.\tag1$$

Such a constraint on $\delta$ is viable by developing a relationship between $\epsilon$ and $\delta$ that looks like
$\displaystyle \delta = \min\left[f(\epsilon), \frac{1}{10}\right]~~$ rather than $~~\displaystyle \delta = f(\epsilon).$

Using (1) above, you have that $~~~\displaystyle 1 - \delta < x < 1 + \delta$.

Then, since $\displaystyle \delta \leq \frac{1}{10} \implies \delta^2 < \delta$
you have that $~~~\displaystyle 1 - 2\delta < x^2 < 1 + 2\delta + \delta^2 < 1 + 3\delta.$
This implies that $~~~\displaystyle 2 - 2\delta < x^2 + 1 < 2 + 3\delta.$

Then, using (1) above, and considering the minimum and maximum possible values of the numerator $(x)$, and denominator $(x^2 + 1)$, you have that

$$\frac{1 - \delta}{2 + 3\delta} < \frac{x}{x^2 + 1} < \frac{1 + \delta}{2 - 2\delta}. \tag2$$

$\displaystyle \frac{1 - \delta}{2 + 3\delta} = \frac{1 + (3/2) \delta}{2 + 3\delta} - \frac{(5/2) \delta}{2 + 3\delta} = \frac{1}{2} - \frac{5\delta}{4 + 6\delta} > \frac{1}{2} - 2\delta ~: ~\delta \leq \frac{1}{10}. $

Similarly, $~~\displaystyle \frac{1 + \delta}{2 - 2\delta} = \frac{1 - \delta}{2 - 2\delta} + \frac{2 \delta}{2 - 2\delta} = \frac{1}{2} + \frac{\delta}{1 - \delta} < \frac{1}{2} + 2\delta ~: ~\delta \leq \frac{1}{10}. $

Therefore, using (2) above,

$$\frac{1}{2} - 2\delta < \frac{x}{x^2 + 1} < \frac{1}{2} + 2\delta \implies - 2\delta < \frac{x}{x^2 + 1} - \frac{1}{2} < 2\delta.$$

This implies that

$$\left|\frac{x}{x^2 + 1} - \frac{1}{2}\right| < 2\delta.\tag3$$

Consequently, set $\displaystyle \delta = \min\left(\frac{\epsilon}{2}, \frac{1}{10}\right).$

Then, based on (1), (2), and (3) above, $~~\displaystyle0 < |x-1| < \delta~~$ will imply that $~~\displaystyle \left|\frac{x}{x^2 + 1} - \frac{1}{2}\right| < \epsilon.$