Argue that there exists a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ which satisfies $g(x) = \frac{1-e^{-2x^2}}{x^2}$ for $x \neq 0$ I have to argue that there exists a continuous  $g: \mathbb{R} \rightarrow \mathbb{R}$ which satisfies that $g(x) = \frac{1-e^{-2x^2}}{x^2}$ for $x \neq 0$ but I am not sure what they actually mean.
Do I just have to argue for that $g(x)$ indeed is continuous as $g(x)$ is a composition of well-known continuous functions? Thus, by properties of continuous functions, $g(x)$ is continuous? 
Furthermore I have to calcualte $g(0)$. However, as $g(x)$ is defined for $x \neq 0$ do I have to find the limit
$$
\lim_{x \rightarrow 0} g(x)
$$
where I can use L'Hopitals rule? 
 A: Since you want $g:\mathbb{R}\to\mathbb{R}$ you want to extend $g$ at $x=0$ in a way that $g$ will still be continous. So we look for $a\in\mathbb{R}$ s.t
$$g(x)=\begin{cases}\dfrac{1-e^{-2x^2}}{x^2},\ x\not=0\\ a,\ x=0\end{cases}$$ is continous. It has to be $\lim\limits_{x\to0}g(x)=g(0)=a$. So we need to calculate $\lim\limits_{x\to0}g(x)$. It is
$$\lim\limits_{x\to0}g(x)=\lim\limits_{x\to0}\dfrac{1-e^{-2x^2}}{x^2}=\lim\limits_{u\to0}\dfrac{1-e^{-2u}}{u}=\lim\limits_{u\to0}\dfrac{(1-e^{-2u})'}{(u)'}=\lim\limits_{u\to0}\dfrac{2e^{-2u}}{1}=2$$
So it has to be $a=2$
A: Let us define
$$g(x) = \begin{cases}\dfrac{1-e^{-2x^2}}{x^2} & x \neq 0\\L & x = 0\end{cases}$$
for some $L \in \Bbb R$ that we will choose later.

It is clear that $g$ is continuous everywhere except possibly at $0$. This is because the composition, addition and quotient of continuous functions is continuous where the denominator is nonzero. This is what you had noted as well.
For $g$ to be continuous, we just have to make sure that it is continuous at $0$. To do this, we simply require
$$\lim_{x\to0}g(x) = g(0).$$
Thus, we shall choose $L$ to be the limit on the left. Again, this is what you had noted. This limit can indeed be calculated using L'Hospital and we get $L = 2$.

Thus, the desired function is
$$g(x) = \begin{cases}\dfrac{1-e^{-2x^2}}{x^2} & x \neq 0\\2 & x = 0\end{cases}$$
This is indeed continuous and sastisfies $g(x) = \dfrac{1-e^{-2x^2}}{x^2}$ for $x \neq 0$.
