Question

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

Short answer

The limit of the vector as \( t \) approaches infinity is \( 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} \) or simply the zero vector.

Step-by-step solution

Evaluate the limit for the i-component

To find the limit as \( t \) approaches infinity for \( e^{-t} \), recognise that as \( t \) increases, \( e^{-t} \) will approach zero because any non-zero number raised to a negative power will get smaller and smaller. Therefore, \( \lim_{t\to\infty} e^{-t} = 0 \). The i-component of the vector becomes \( 0\mathbf{i} \).

Evaluate the limit for the j-component

To find the limit as \( t \) approaches infinity for \( \frac{1}{t} \), realise that as \( t \) gets larger and larger, \( \frac{1}{t} \) will approach zero because the fraction's denominator becomes larger and larger. Therefore, \( \lim_{t\to\infty} \frac{1}{t} = 0 \). The j-component of the vector becomes \( 0\mathbf{j} \).

Evaluate the limit for the k-component

To find the limit as \( t \) approaches infinity for \( \frac{t}{t^2+1} \), apply L'Hospital's Rule, which states that for the limit \( \lim_{t\to\infty} \frac{f(t)}{g(t)} \), where both \( f(t) \) and \( g(t) \) approach infinity, the limit is equivalent to \( \lim_{t\to\infty} \frac{f'(t)}{g'(t)} \). So differentiate the numerator to get \( 1 \) and the denominator to get \( 2t \). The limit therefore becomes \( \lim_{t\to\infty} \frac{1}{2t} \), which is zero. So, the k-component of the vector becomes \( 0\mathbf{k} \).