Chapter 14: Problem 35
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
Expert verified
Question: What is the limit as t approaches 0 of the given vector:
$$ \lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^t-t-1}{t}\mathbf{j}+\frac{\cos t+\frac{t^2}{2}-1}{t^2} \mathbf{k}\right) $$
Answer: The limit as t approaches 0 of the given vector is $\mathbf{i}$.
Step by step solution
01
Calculate the limit for the i-component
To evaluate the limit for \(t\) approaching \(0\) of the expression related to the unit vector \(\mathbf{i}\), we will use the standard limit:
$$ \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 $$
Thus, the limit for the i-component is:
$$ \lim_{t \rightarrow 0} \frac{\sin t}{t} = 1 $$
02
Calculate the limit for the j-component
To evaluate the limit for \(t\) approaching \(0\) of the expression related to the unit vector \(\mathbf{j}\), we will use L'Hôpital's Rule since we have a \(\frac{0}{0}\) indeterminate form:
$$ \lim_{t\rightarrow 0}\frac{e^t - t - 1}{t} $$
Differentiate the numerator and denominator with respect to \(t\), and we get:
$$ \lim_{t \rightarrow 0} \frac{e^t - 1}{1} $$
Now, applying the limit, we have:
$$ \lim_{t \rightarrow 0} \frac{e^0 - 1}{1} = \frac{1 - 1}{1} = 0 $$
03
Calculate the limit for the k-component
To evaluate the limit for \(t\) approaching \(0\) of the expression related to the unit vector \(\mathbf{k}\), we will again use L'Hôpital's Rule since we have a \(\frac{0}{0}\) indeterminate form:
$$ \lim_{t \rightarrow 0} \frac{\cos t + \frac{t^2}{2} - 1}{t^2} $$
Differentiate the numerator and denominator with respect to \(t\), and we get:
$$ \lim_{t \rightarrow 0} \frac{-\sin t + t}{2t} $$
Now, we still have a \(\frac{0}{0}\) indeterminate form, so we apply L'Hôpital's Rule once again, differentiating the numerator and denominator with respect to \(t\):
$$ \lim_{t \rightarrow 0} \frac{-\cos t + 1}{2} $$
Applying the limit, we have:
$$ \lim_{t \rightarrow 0} \frac{-\cos 0 + 1}{2} = \frac{-1 + 1}{2} = 0 $$
04
Combine the components
Now, combine the limits of each component to find the overall limit for the given vector:
$$ \lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^t-t-1}{t}\mathbf{j}+\frac{\cos t+\frac{t^2}{2}-1}{t^2} \mathbf{k}\right) = (1\mathbf{i} - 0\mathbf{j} + 0\mathbf{k}) = \mathbf{i}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits of Vector-Valued Functions
In multivariable calculus, understanding limits of vector-valued functions is essential. These functions can be a bit more complex than those with real values, as they involve vectors which have both magnitude and direction. A vector-valued function takes the form \( \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} \) where \( f(t) \), \( g(t) \), and \( h(t) \) are component functions, and \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) are unit vectors along the coordinate axes.To evaluate the limit of such functions as \( t \) approaches a certain value, you simply need to consider the limit of each component function separately:
- i-component: \( \lim_{t \to c} f(t)\mathbf{i} \)
- j-component: \( \lim_{t \to c} g(t)\mathbf{j} \)
- k-component: \( \lim_{t \to c} h(t)\mathbf{k} \)
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus, especially useful for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule helps in simplifying functions into a more manageable form by differentiating the numerator and the denominator.
When to Use L'Hôpital's Rule
You can apply this rule when:- The function \( \frac{f(t)}{g(t)} \) results in an indeterminate form.
- Both \( f(t) \) and \( g(t) \) are differentiable near the point of interest.
- The derivative \( \frac{f'(t)}{g'(t)} \) exists.
Applying the Rule
For the j and k-components of the given exercise, L'Hôpital's Rule was applied to handle the indeterminate forms:- Differentiate the numerator and denominator separately.
- Re-evaluate the limit of the new form.
Unit Vectors
Unit vectors are fundamental components in the study of vectors in calculus and physics. A unit vector has a magnitude of 1 and provides direction along a particular axis in a coordinate system.
Common Unit Vectors
In three-dimensional space, the common unit vectors are:- \( \mathbf{i} \): Shows direction along the x-axis.
- \( \mathbf{j} \): Shows direction along the y-axis.
- \( \mathbf{k} \): Shows direction along the z-axis.