Evaluate the following limit. \[ \lim _{t \rightarrow \infty}\left(e^{-t} i-\frac{2 t}{t+7} j+5 \tan ^{-1} t k\right) \] \[ \lim _{t \rightarrow \infty}\left(e^{-t} i-\frac{2 t}{t+7} j+5 \tan ^{-1} t k\right)= \] (Type an exact answer, using رas needed.)

To evaluate the limit \[ \lim _{t \rightarrow \infty}\left(e^{-t} i-\frac{2 t}{t+7} j+5 \tan ^{-1} t k\right), \] we analyze each component separately as \(t\) approaches infinity. 1. **For the first term**: \(e^{-t} i\). - As \(t \rightarrow \infty\), \(e^{-t} \rightarrow 0\). Thus, the limit of this term is \(0i = 0\). 2. **For the second term**: \(-\frac{2 t}{t+7} j\). - We can simplify \(\frac{2 t}{t+7}\): \[ \frac{2 t}{t+7} = \frac{2 t}{t(1 + \frac{7}{t})} = \frac{2}{1 + \frac{7}{t}}. \] - As \(t \rightarrow \infty\), \(\frac{7}{t} \rightarrow 0\), so \(\frac{2 t}{t+7} \rightarrow \frac{2}{1} = 2\). - Thus, \(-\frac{2 t}{t+7} j \rightarrow -2j\). 3. **For the third term**: \(5 \tan^{-1} t k\). - As \(t \rightarrow \infty\), \(\tan^{-1} t \rightarrow \frac{\pi}{2}\). Therefore, \(5 \tan^{-1} t \rightarrow 5 \cdot \frac{\pi}{2} = \frac{5\pi}{2}\). - Thus, \(5 \tan^{-1} t k \rightarrow \frac{5\pi}{2} k\). Putting all these together, we have: \[ \lim _{t \rightarrow \infty}\left(e^{-t} i-\frac{2 t}{t+7} j+5 \tan ^{-1} t k\right) = 0i - 2j + \frac{5\pi}{2}k. \] Therefore, the result is: \[ -\mathbf{2} j + \frac{5\pi}{2} k. \] So the final answer is: \[ \boxed{-2 j + \frac{5\pi}{2} k}. \]