Question

Evaluate the following limits. $$\lim _{t \rightarrow \infty}\left(e^{-t} \mathbf{i}-\frac{2 t}{t+1} \mathbf{j}+\tan ^{-1} t \mathbf{k}\right)$$

Short answer

Answer: The limit of the vector as t approaches infinity is $$0\mathbf{i} - 2\mathbf{j} + \frac{\pi}{2}\mathbf{k}$$.

Step-by-step solution

i Component

To evaluate the i-component, we only need to look at the exponential part of the vector: $$\lim_{t\rightarrow\infty}(e^{-t}\mathbf{i})$$. Using the property of limits for exponentials, we know that as t approaches infinity: $$e^{-t} \rightarrow 0$$ Thus, the limit of the i-component is: $$0\mathbf{i}$$.

j Component

The j-component is given by the following expression: $$ \lim_{t\rightarrow\infty}\left(-\frac{2t}{t+1}\mathbf{j}\right) $$. To find the limit, we use L'Hopital's rule for indeterminate forms (0/0 or infinity/infinity). Differentiate both the numerator and denominator with respect to t: Numerator: $$\frac{d}{dt}(2t) = 2$$ Denominator: $$\frac{d}{dt}(t+1) = 1$$ Now, divide the derivatives: $$-\frac{2}{1}$$ As t approaches infinity, there's no more 't' term left in the expression, therefore, the limit of the j-component is: $$-2\mathbf{j}$$.

k Component

For the k-component, we need to evaluate the limit: $$\lim_{t\rightarrow\infty}(\tan^{-1}(t)\mathbf{k})$$ Recall that the inverse tangent function has horizontal asymptotes at $$\frac{\pi}{2}$$ and $$-\frac{\pi}{2}$$. As t approaches infinity, the value of the inverse tangent function approaches the horizontal asymptote at $$\frac{\pi}{2}$$. Therefore, the limit of the k-component is: $$\frac{\pi}{2}\mathbf{k}$$.

Combining the Components

Now that we found the limits for each component, i, j, and k, we can combine them to get the overall vector limit. This gives us: $$\lim_{t\rightarrow\infty}\left(e^{-t}\mathbf{i}-\frac{2t}{t+1}\mathbf{j}+\tan^{-1}t\mathbf{k}\right) = 0\mathbf{i} - 2\mathbf{j} + \frac{\pi}{2}\mathbf{k}$$