Consider the function $f(x)=\frac{1}{x^3}$. I am looking for a proper cut of $f$ which returns a function which removes the singularity in $0$ and remains continuous in $\mathbb{R}$.
Could someone please suggest me something?
I was thinking about something like a function which is equal to $f$ outside the ball $B_{\varepsilon}(0)$, $\varepsilon>0$, but I don't know how to assume it inside the ball.
Thank you in advance!
Assume that we want to change the function only in $\{|x| \leq \varepsilon\}$, and to it in the simplest way possible -- connect the end points with a line. The line would have to pass through two end points which are $$ (x_0,y_0) = \left(\varepsilon, \tfrac{1}{\varepsilon^3}\right) \quad \text{ and }\quad (x_1,y_1) = \left(-\varepsilon, \tfrac{-1}{\varepsilon^3}\right) $$
The line passing through $(x_0,y_0)$ and $(x_1,y_1)$ is $y=x\cdot \frac{1}{\varepsilon^4}$. Therefore the function is $$ f(x) = \begin{cases} \frac{1}{x^3} &\text{if $|x|\geq \varepsilon$}, \\ x \frac{1}{\varepsilon^4} &\text{ if $|x| < \varepsilon$}. \end{cases} $$