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Chapter 11: Problem 46

Evaluate the following limits. $$\lim _{t \rightarrow 0}\left(\frac{\tan t}{t} \mathbf{i}-\frac{3 t}{\sin t} \mathbf{j}+\sqrt{t+1} \mathbf{k}\right)$$

Short Answer

Expert verified
Question: Evaluate the following limit as \(t \rightarrow 0\): $$\lim_{t \rightarrow 0} \left(\frac{\tan t}{t}\mathbf{i} - \frac{3t}{\sin t}\mathbf{j} + \sqrt{t+1}\mathbf{k}\right)$$ Answer: The given limit evaluates to \(\mathbf{1i - 3j + k}\).

Step by step solution

01

Evaluate the limit of the first component

To evaluate the limit for the first component, \(\frac{\tan t}{t}\), as \(t \rightarrow 0\), we can use l'Hopital's rule since it satisfies the indeterminate form \(\frac{0}{0}\). Taking the derivative of the numerator and denominator, and then finding the limit, we get: $$\lim _{t \rightarrow 0}\left(\frac{\tan t}{t}\right) = \lim _{t \rightarrow 0}\left(\frac{\sec^2 t}{1}\right) = \frac{\sec^2 0}{1} = \frac{1}{1} = 1$$
02

Evaluate the limit of the second component

To evaluate the limit for the second component, \(\frac{3 t}{\sin t}\), as \(t \rightarrow 0\), we can again use l'Hopital's rule as it also satisfies the indeterminate form \(\frac{0}{0}\). Taking the derivative of the numerator and denominator, and then finding the limit, we get: $$\lim _{t \rightarrow 0}\left(\frac{3 t}{\sin t}\right) = \lim _{t \rightarrow 0}\left(\frac{3}{\cos t}\right) = \frac{3}{\cos 0} = \frac{3}{1} = 3$$
03

Evaluate the limit of the third component

To evaluate the limit for the third component, \(\sqrt{t+1}\), as \(t \rightarrow 0\), we can directly substitute the value of \(t\) since it does not form any indeterminate form. So we get: $$\lim _{t \rightarrow 0}\left(\sqrt{t+1}\right) = \sqrt{0+1} = \sqrt{1} = 1$$
04

Combine the limits of the components

Now that we have the limits for each component, we can combine them to get the overall limit of the given expression: $$\lim _{t \rightarrow 0}\left(\frac{\tan t}{t} \mathbf{i}-\frac{3 t}{\sin t} \mathbf{j}+\sqrt{t+1} \mathbf{k}\right) = \left(1 \mathbf{i} - 3 \mathbf{j} + 1 \mathbf{k}\right)$$

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