I need to approximate function $f(x) = \frac{1}{x}$, using an exponential function like $g(x)$, using a parameter,$\sigma$ which when $\sigma$ goes to 0, g(x) becomes $g(x)$ equal to $f(x)$. Like $g(x) = exp(-x -\sigma)$, but it doesn't equal to $f(x)$ when $\sigma=0$. Is there any such function?
Edit: I want a function which doesn't have any term like $\frac{1}{x}$. Otherwise, I work with the original function. It is best to have summation of exponential.
If I understood you correctly, you want a function $g(x,\sigma)$, where $\sigma$ is a tuning parameter, than controls how close $g$ is to $\frac{1}{x}$; when $\sigma=0$, we want the difference to be zero. Let $$g(x,\sigma)=\frac{e^{\frac{\sigma}{x}}-1}{\sigma}=\sum_{k=1}^\infty \frac{\sigma^{k-1}}{k!x^k}$$ Than $$\lim_{\sigma\to 0}g(x,\sigma)=\frac{1}{x}$$ as requested. I hope I understood your request correctly.
If you want your function to contain only exponential terms, we can define
$$g(x,\sigma)=\frac{e^{\frac{\sigma}{x}}-e^{x\sigma^2}}{\sigma}$$
To achieve the same limit.
Edit
After the clarifying edit to the original question, this answer is not viable.