91Ó°ÊÓ

To find the limit of this component, we can use L'Hôpital's Rule, which states that: $$\lim_{t \rightarrow 0} \frac{f(t)}{g(t)} = \lim_{t \rightarrow 0} \frac{f'(t)}{g'(t)}$$ when the limit of the numerator and denominator are 0. In this case, \(f(t) = \sin t\) and \(g(t) = t\). Taking the derivative of both, we get: $$f'(t) = \cos t$$ $$g'(t) = 1$$ Applying L'Hôpital's Rule, we have: $$\lim_{t \rightarrow 0} \frac{\sin t}{t} = \lim_{t \rightarrow 0} \frac{\cos t}{1} = \frac{\cos 0}{1} = 1$$

Again, we use L'Hôpital's Rule. This time, \(f(t) = e^{t} - t - 1\) and \(g(t) = t\). Taking the derivative of both, we get: $$f'(t) = e^{t} - 1$$ $$g'(t) = 1$$ Applying L'Hôpital's Rule, we have: $$\lim_{t \rightarrow 0} \frac{e^{t} - t - 1}{t} = \lim_{t \rightarrow 0} \frac{e^{t} - 1}{1} = \frac{e^{0} - 1}{1} = 0$$

Since both the numerator and denominator tend to 0 when \(t\rightarrow0\), L'Hôpital's Rule can be applied here as well. The first derivatives of the numerator and denominator are: $$f'(t) = -\sin t + t$$ $$g'(t) = 2t$$ Applying L'Hôpital's Rule for the first time gives: $$\lim_{t \rightarrow 0} \frac{\cos t + \frac{t^{2}}{2} - 1}{t^{2}} = \lim_{t \rightarrow 0} \frac{-\sin t + t}{2t}$$ However, to reach the limit, we must apply L'Hôpital's Rule again since both numerator and denominator still tend to 0 as \(t\rightarrow0\). We find the second derivatives of the numerator and denominator: $$f''(t) = -\cos t + 1$$ $$g''(t) = 2$$ Applying L'Hôpital's Rule for the second time, we have: $$\lim_{t \rightarrow 0} \frac{-\sin t + t}{2t} = \lim_{t \rightarrow 0} \frac{-\cos t + 1}{2} = \frac{-\cos 0 + 1}{2} = \frac{0}{2} = 0$$

We can now combine the limits of the individual components to find the overall limit as \(t \rightarrow 0\): $$\lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^{t}-t-1}{t} \mathbf{j}+\frac{\cos t+t^{2} / 2-1}{t^{2}} \mathbf{k}\right) = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$$