<s>[INST] 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!
 [/INST] 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:
$$
\begin{align}
|f(x)-f(y)|&=\left|\frac{x-y}{K}+\frac{xy}{k^2}(y-x)\right|\\
&\leq \left|\frac{x-y}{K}\right|+\left|\frac{xy}{K^2}(y-x)\right|\\
&=\frac{\left|x-y\right|}{\left|K\right|}\left|1-\frac{xy}{K}\right|\\
&\leq\frac{\left|x-y\right|}{\left|K\right|}\left(1+\frac{\left|xy\right|}{\left|K\right|}\right)\\
&<\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\left|xy\right|}{\left|K\right|}\right)\\
&<\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\left|\delta(\epsilon,y)y+y^2\right|}{\left|K\right|}\right)\\
&\leq\frac{\delta(\epsilon,y)}{\left|K\right|}\left(1+\frac{\delta(\epsilon,y)|y|+y^2}{\left|K\right|}\right)\\
&=\delta^2(\epsilon,y)\frac{y}{K^2}+\delta(\epsilon,y)\left(\frac{y^2}{K^2}+\frac{1}{\left|K\right|}\right)\\
&=\epsilon \\
\end{align}
$$
</s>