If $f\in C^{\infty}(\mathbb{R})$ and $f(0)=0,$ let $g(t)=\frac{f(t)}{t},$ will $g$ be in $C^{\infty}(\mathbb{R})$?

Question

Suppose $f\in C^{\infty}(\mathbb{R})$ satisfies $f(0)=0$ (and $f^{\prime}(0)\ne 0,$ I don't know whether this condition is necessary or not so I put it in parentheses), define $g:\mathbb{R}\to\mathbb{R}$ to be $$g(t)=\begin{cases}f(t)/t, & t\ne 0 \\ f^{\prime}(0), & t=0\end{cases}$$ by L'Hospital's law we see $g\in C^0(\mathbb{R})\cap C^{\infty}(\mathbb{R}\backslash \left\{0\right\}),$ now I want to obtain the smoothness of $g$ at $t=0,$ the only way I come about is to differentiate it using Leibniz's rule for $t\ne 0$ and use definition of derivative at $t=0,$ but the result seems complicated and I cannot catch it, can someone help me?