Answer
$$\left\langle 1 ,1 ,0\right\rangle $$
Work Step by Step
By making use of L'Hôpital's rule on the second component, we have
$$
\lim _{t \rightarrow 0} \left\langle \frac{1}{t+1} , \frac{e^t-1}{t} ,4t\right\rangle=\\ \left\langle\lim _{t \rightarrow 0}\frac{1}{t+1} ,\lim _{t \rightarrow 0} \frac{e^t-1}{t} ,\lim _{t \rightarrow 0}4t\right\rangle\\
=\left\langle 1 ,1 ,0\right\rangle
$$