Show a function is continuous on $\mathbb{R}$ by any method

Question

Let $f(x)=\frac{Kx}{K^2+x^2}$ where K is some constant, show this is continuous on $\mathbb{R}$.

Here are my scratch work in looking for a delta. let $x,y\in \mathbb{R} $ WTS $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|<\epsilon,\forall \epsilon>0$ whenever $|x-y|<\delta$. So, $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|=|K||\frac{x(K^2+y^2)-y(K^2+x^2)}{(K^2+y^2)(K^2+x^2)}|\leq |\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}|\leq|\frac{xK^2-yK^2+xy^2}{(K^2+y^2)(K^2+x^2)}|$ I'm a bit stuck here on how to relate this inequality to $|x-y|$. Can someone help out? Thanks!

Answer 1

Though you agreed to use a more appropriate general argument (see comments), let us finish your $\epsilon-\delta$-proof (assuming $K\ne0$).

You were stuck on how to relate the inequality $$\left|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}\right|\le|K|\left|\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}\right|$$ to $|x−y|.$ Just factorize it in the numerator: $xK^2-yK^2+xy^2-yx^2=(x-y)(K^2-xy)$ hence for $|x-y|<\delta$, $$\left|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}\right|\le\delta|K|M\quad\text{where}\quad M:=\frac{K^2+|xy|}{(K^2+y^2)(K^2+x^2)}$$ and you just need to bound $M$. If you only want to prove the continuity at any fixed point $x$, you can end up like this: wlog $\delta<1$, so that $$M\le\frac{K^2+|x|(|x|+1)}{K^4}.$$ But you can even prove that $f$ is uniformly continuous: $$M\le\frac{K^2}{K^4}+\frac{|xy|}{K^2(x^2+y^2)}\le\frac1{K^2}+\frac1{2K^2}.$$

Answer 2

Since you are asking a solution by any method, you can try to use the limit definition of continuity. Recall that a function $f$ is continuous at a point $c$ of its domain if $$\lim_{x\rightarrow c} f(x) = f(c)$$

So for any $c\in \mathbb{R}$ we have $\lim_{x\rightarrow c}Kx = Kc$ and $\lim_{x\rightarrow c} K^2+x^2 = K^2+c^2 \neq 0$. So we have:$$ f(c) = \frac{Kc}{K^2+c^2} = \frac{\lim_{x\rightarrow c}Kx}{\lim_{x\rightarrow c}K^2+c^2} = \lim_{x\rightarrow c}\frac{Kx}{K^2+x^2} = \lim_{x\rightarrow c} f(x) $$

Answer 3

Since you started your proof attempting to use the $\epsilon$-$\delta$ definition here is a hint. One can proceed in the following way:

1. If $K=0$, then $f(x)=0$ for all $x$ and the proof is trivial.

2. If $K\neq 0$, then we can express:

$$f(x)=\frac{x/K}{1+x^2/K^2}$$.

The $\epsilon$-$\delta$ proof is a bit laborious, but my best guess is that $$\delta(\epsilon,y)=\frac{-(\frac{y^2}{K^2}+\frac{1}{{|K|}})+\sqrt{\frac{y^4}{K^4}+\frac{2y^2}{{|K|}^3}+\frac{1+4\epsilon y}{K^2}}}{2|y|/K^2}$$

should do the trick. Do you see why?

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Details:

Let $|x-y|<\delta(\epsilon,y)$. Then:

$$ \begin{align} |f(x)-f(y)|&=\left|\frac{x-y}{K}+\frac{xy}{k^2}(y-x)\right|\\ &\leq \left|\frac{x-y}{K}\right|+\left|\frac{xy}{K^2}(y-x)\right|\\ &=\frac{\left|x-y\right|}{\left|K\right|}\left|1-\frac{xy}{K}\right|\\ &\leq\frac{\left|x-y\right|}{\left|K\right|}\left(1+\frac{\left|xy\right|}{\left|K\right|}\right)\\ &<\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\left|xy\right|}{\left|K\right|}\right)\\ &<\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\left|\delta(\epsilon,y)y+y^2\right|}{\left|K\right|}\right)\\ &\leq\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\delta(\epsilon,y)|y|+y^2}{\left|K\right|}\right)\\ &=\delta^2(\epsilon,y)\frac{y}{K^2}+\delta(\epsilon,y)\left(\frac{y^2}{K^2}+\frac{1}{\left|K\right|}\right)\\ &=\epsilon \\ \end{align} $$