Show a function is continuous on $\mathbb{R}$ by any method 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!
 A: 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:
$$M\le\frac{K^2}{K^4}+\frac{|xy|}{K^2(x^2+y^2)}\le\frac1{K^2}+\frac1{2K^2}.$$
A: 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:$$ 
    f(c) = \frac{Kc}{K^2+c^2} = \frac{\lim_{x\rightarrow c}Kx}{\lim_{x\rightarrow c}K^2+c^2} = \lim_{x\rightarrow c}\frac{Kx}{K^2+x^2} = \lim_{x\rightarrow c} f(x)
$$
A: Since you started your proof attempting to use the $\epsilon$-$\delta$ definition here is a hint. One can proceed in the following way:

*

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


*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?

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