Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 4

$$\left\langle 1 ,1 ,0\right\rangle$$

Work Step by Step

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

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