Infinite limit of an infinite sequence

Question

Given the following infinite series:

$$f(x) =\sum_{n=1}^\infty \frac{(-1)^{n-1} x^{2n-1}}{(n-1)!(2n-1)} = x - \frac{x^3}{1!\cdot 3}+\frac{x^5}{2!\cdot 5} - \frac{x^7}{3!\cdot 7}+ \dots$$

How does one take the limit of of such a series as $x$ tends to infinity?

That is, how do you calculate: $\lim_{x\to \infty} f(x)$

According to WolframAlpha, the answer is $\frac{\sqrt{\pi}}{2}$. Their solution, however, involves some error function they defined which is foreign to me. Is there any other way to compute this?

Answer 1

Note that $f'(x)=e^{-x^2}$, hence $$\lim_{x\to \infty} f(x)=\lim_{x\to \infty} (f(x)-f(0))=\int_0^{\infty}f'(x) dx =\int_0^{\infty}e^{-x^2} dx=\frac{\sqrt{\pi}}{2}.$$ As regards the last step you can take a look here: Is there really no way to integrate $e^{-x^2}$?