How to prove this limit using epsilon delta method?

Question

i know how to construct a proof using epsilon delta method but as much as i have done i have to just algebraically manipulate and i got the value but i cant get this one.

$$\lim_{x \to c} \frac{x}{1+x^2} = \frac{c}{1+c^2}$$

My try: $$\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|<\epsilon\implies\left|\frac{x^2-c^2}{1+c^2+x^2+x^2c^2}\right|<\epsilon\implies\left|\frac{(x-c)(x+c)}{1+c^2+x^2+x^2c^2}\right|<\epsilon$$

But i dont how to proceed after this?

Answer 1

One has $${x\over 1+x^2}-{c\over 1+c^2}=(x-c){1-cx\over(1+x^2)(1+c^2)}$$ and therefore $$\left|{x\over 1+x^2}-{c\over 1+c^2}\right|\leq|x-c|\bigl(1+|c|\>|x|\bigr)\ .\tag{1}$$ When $|x-c|\leq2$ then $|x|\leq |c|+2$ and therefore $$1+|c|\>|x|\leq\bigl(1+|c|\bigr)^2\ ,\tag{2}$$ independently of $x$.

Now let an $\epsilon>0$ be given. I claim that $$\delta:=\min\left\{{\epsilon\over (1+|c|)^2}, \ 2\right\}\tag{3}$$ does the job. Here the "$2$" in $(3)$ is only a safety measure to ensure that in any case the $\delta$ so defined will ensure $|x-c|\leq2$, so that $(2)$ is valid.

Proof. Assume that $|x-c|<\delta$. Then from $(1)$ and $(2)$ we obtain $$\left|{x\over 1+x^2}-{c\over 1+c^2}\right|\leq|x-c|\bigl(1+|c|\>|x|\bigr)<\delta\bigl(1+|c|\bigr)^2\leq\epsilon\ .$$

Answer 2

$$\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|=\left|\frac{(x-c)(1-xc)}{(1+x^2)(1+c^2)} \right|\le \left|\frac{1+|x||c|}{(1+x^2)(1+c^2)} \right||x-c|$$ Suppose $c\not=0$, then we do some analysis first.

If we require $|x-c|<\frac{|c|}{2}$, then $|1+x^2|\ge 1+\frac{c^2}{4},|x|<\frac{3|c|}{2}$, that is $\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|\le M(c)|x-c|$, where $M(c)>0$ is a fixed constant. Hence it suffices to further choose $|x-c|<\frac{\epsilon}{M(c)}$.

Combine above two parts together, let $\delta=\min(\frac{|c|}{2},\frac{\epsilon}{M(c)})$, and then $|x-c|<\delta\implies \left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|<\epsilon$

If $c=0$, you are proving $\lim_{x\to 0}\frac{x}{1+x^2}=0$, which is easier by squeezing theorem.

Answer 3

If $c=0,$ take $\delta = \epsilon.$ If $c\ne 0,$ take $\delta = \min (\epsilon/(1+2c^2),|c|).$ Why? We have

$$\left |\frac{(x-c)(1-xc)}{(1+x^2)(1+c^2)} \right | \le |x-c|\,|1-xc|.$$

If $c=0,$ then we're left with $|x-c|$ on the right, perfect. If $c\ne 0,$ $|x-c|<|c| \implies |x|\le |x-c|+|c|<2|c|\implies |1-xc| < 1+2c^2.$ As $|x-c|<\epsilon/(1+2c^2),$ we're done.

Answer 4

Letting $f(x) = \frac x{1+x^2}$ and taking the derivative, I see that $|f'(x)| \leq 1$ everywhere. This suggests that something like $\delta = \epsilon$ might work.

I start with the calculation

$$\left|\frac{x}{1+x^2} - \frac{c}{1+c^2}\right| = \left|\frac{(x-c)(1-xc)}{(1+x^2)(1+c^2)}\right| = \left|\frac 1{1+x^2}\right| \left|\frac{1-xc}{1+c^2} \right| |x-c|.$$

The part after the second $=$ sign is broken up into factors that I think I might be able to bound. For any $x$, $\left|\frac 1{1+x^2}\right| \leq 1,$ so

$$\left|\frac{x}{1+x^2} - \frac{c}{1+c^2}\right| < \left|\frac{1-xc}{1+c^2} \right| |x-c|. \tag{1}$$

If I let $0 < |x - c| < \delta = \epsilon$, if I can just show that $\left|\frac{1-xc}{1+c^2} \right| \leq 1$ then I am done. But there is a slight hitch; for some values of $\epsilon$ I could satisfy $0 < |x - c| < \epsilon$ by setting $x = -2c$, and $\left|\frac{1-(-2c)c}{1+c^2} \right| = \left|\frac{1+2c^2}{1+c^2} \right| > 1$ whenever $c \neq 0$. This only happens for relatively large values of $\epsilon$, but we must cover all values.

So it turns out that $\delta = \epsilon$ does not always work. But if I also stipulate that $|x - c| < |c|$ when $c \neq 0$, then $|x| < 2|c|$, so $|1 - xc| \leq 1 + 2c^2$ and $\left|\frac{1-xc}{1+c^2} \right| \leq 2$. That's a factor of $2$ larger than I wanted, but it's easy to compensate for that factor by requiring $|x - c| < \frac\epsilon2$ instead of just $|x - c| < \epsilon$.

Putting it all together, I have to consider just two cases:

• If $c=0$ then I set $\delta = \epsilon$.
• If $c\neq 0$ then I set $\delta = \min\left(\frac\epsilon2, |c|\right)$.

In either case I find that the right-hand side of inequality $(1)$ is less than $\epsilon$, so $|f(x) - f(c)| < \epsilon.$

The key to my thinking here is that this is not a piece of precision machinery. It is more like splitting logs for firewood. The results do not have to precisely conform to an exact predetermined shape and size; it is sufficient that they be small enough. So I keep on chopping until they are.