Question

Evaluate the following limits. $$\lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^{t}-t-1}{t} \mathbf{j}+\frac{\cos t+t^{2} / 2-1}{t^{2}} \mathbf{k}\right)$$

Short answer

Answer: The limit is $\mathbf{i}$.

Step-by-step solution

Find i-component limit

To find the limit of the i component, we need to evaluate the following limit: $$ \lim_{t\to0}\frac{\sin t}{t} $$ This is a well-known limit, which is equal to 1. So, the i component is: $$ \mathbf{i}(1) $$

Find j-component limit

To find the limit of the j component, we need to evaluate the following limit: $$ \lim_{t\to0}\frac{e^t - t - 1}{t} $$ By applying the L'Hôpital's rule, we differentiate the numerator and the denominator with respect to t and get the new limit: $$ \lim_{t\to0}\frac{e^t - 1}{1} $$ Now, we plug in t=0 to get the value of the limit, which is: $$ \frac{e^0 - 1}{1} = 0 $$ So, the j component is: $$ \mathbf{j}(0) $$

Find k-component limit

To find the limit of the k component, we need to evaluate the following limit: $$ \lim_{t\to0}\frac{\cos t + \frac{t^2}{2} - 1}{t^2} $$ By applying the L'Hôpital's rule two times, we differentiate the numerator and denominator with respect to t and get the new limits: $$ \lim_{t\to0}\frac{-\sin t + t}{2t} = \lim_{t\to0}\frac{-\cos t + 1}{2} $$ Now, we plug in t=0 to get the value of the limit, which is: $$ \frac{-\cos 0 + 1}{2} = \frac{-1+1}{2}=0 $$ So, the k component is: $$ \mathbf{k}(0) $$

Combine the components

Now, combine the calculated components to find the limit of the given vector expression: $$ \lim_{t\rightarrow 0}\left(\frac{\sin t}{t}\mathbf{i}-\frac{e^{t}-t-1}{t}\mathbf{j}+\frac{\cos t+t^{2}/2-1}{t^2}\mathbf{k}\right) = \mathbf{i}(1) + \mathbf{j}(0) + \mathbf{k}(0) = \boxed{\mathbf{i}} $$