Question

Find \(\lim _{r \rightarrow \infty} \mathbf{X}_{r}\) (a) \(\mathbf{X}_{r}=\left(r \sin \frac{\pi}{r}, \cos \frac{\pi}{r}, e^{-r}\right)\) (b) \(\mathbf{X}_{r}=\left(1-\frac{1}{r^{2}}, \log \frac{r+1}{r+2},\left(1+\frac{1}{r}\right)^{r}\right)\)

Short answer

(a) \(\mathbf{X}_{r}=\left(r \sin \frac{\pi}{r}, \cos \frac{\pi}{r}, e^{-r}\right)\) (b) \(\mathbf{X}_{r}=\left(1-\frac{1}{r^{2}}, \log \frac{r+1}{r+2},\left(1+\frac{1}{r}\right)^{r}\right)\) Answer: (a) The limit of the vector function \(\mathbf{X}_{r}\) as \(r\) approaches infinity is \(\left(-\pi, 1, 0\right)\). (b) The limit of the vector function \(\mathbf{X}_{r}\) as \(r\) approaches infinity is \(\left(1, 0, e\right)\).

Step-by-step solution

Compute the limit of each element

Compute the limit of each element in the vector as \(r \rightarrow \infty\) separately: 1. \(\lim_{r \rightarrow \infty} r \sin \frac{\pi}{r}\) 2. \(\lim_{r \rightarrow \infty} \cos \frac{\pi}{r}\) 3. \(\lim_{r \rightarrow \infty} e^{-r}\)

Use L'Hopital's rule for the first limit

The first limit has an indeterminate form "infinity times 0". To solve this, we need to rewrite it as a quotient of two functions: \(\lim_{r \rightarrow \infty} r \sin \frac{\pi}{r} = \lim_{r \rightarrow \infty} \frac{\sin \frac{\pi}{r}}{\frac{1}{r}}\) Now, using L'Hopital's rule and taking derivative of numerator and the denominator will give: Result = \(\lim_{r \rightarrow \infty} \frac{\frac{\pi}{r^2} \cos \frac{\pi}{r}}{-\frac{1}{r^2}} = -\pi\)

Compute the remaining limits

We can now compute the remaining limits: 2. \(\lim_{r \rightarrow \infty} \cos \frac{\pi}{r} = \cos 0 = 1\) 3. \(\lim_{r \rightarrow \infty} e^{-r} = 0\)

Write the final result for part (a)

Combine the limits from each element to get the final result: \(\lim_{r \rightarrow \infty} \mathbf{X}_{r} = \left(-\pi, 1, 0\right)\) #(b) Find the limit of \(\mathbf{X}_{r}\) as r approaches infinity# (b) \(\mathbf{X}_{r}=\left(1-\frac{1}{r^{2}}, \log \frac{r+1}{r+2},\left(1+\frac{1}{r}\right)^{r}\right)\)

Compute the limit of each element

Compute the limit of each element in the vector as \(r \rightarrow \infty\) separately: 1. \(\lim_{r \rightarrow \infty} 1-\frac{1}{r^{2}}\) 2. \(\lim_{r \rightarrow \infty} \log \frac{r+1}{r+2}\) 3. \(\lim_{r \rightarrow \infty} \left(1+\frac{1}{r}\right)^{r}\)

Compute the first and second limit

The first and second limits are straightforward to compute: 1. \(\lim_{r \rightarrow \infty} 1-\frac{1}{r^{2}} = 1\) 2. \(\lim_{r \rightarrow \infty} \log \frac{r+1}{r+2} = \log 1 = 0\)

Compute the third limit

The third limit is a well-known limit: 3. \(\lim_{r \rightarrow \infty} \left(1+\frac{1}{r}\right)^{r} = e\)

Write the final result for part (b)

Combine the limits from each element to get the final result: \(\lim_{r \rightarrow \infty} \mathbf{X}_{r} = \left(1, 0, e\right)\)

Key concepts

Limit of a Vector

Understanding the limit of a vector involves deconstructing the vector into its individual components and finding the limit of each component separately. Similar to how we approach the limit of a single-variable function, the limit of a vector-valued function as a variable, say r, approaches infinity, is found by considering the behavior of each component of the vector as r increases without bound.

In the given exercise, we are provided with vectors in multi-dimensional space and are asked to compute the limit for each coordinate as r approaches infinity. For the vector \( \mathbf{X}_{r}=(r \sin \frac{\pi}{r}, \cos \frac{\pi}{r}, e^{-r}) \), the components converge to definite values, which gives us the limit of the entire vector. Likewise, for the vector \( \mathbf{X}_{r}=(1-\frac{1}{r^{2}}, \log \frac{r+1}{r+2},\left(1+\frac{1}{r}\right)^{r}) \), each component has a clear limit as r progressively gets larger. These exercises refine our understanding of vector limits by decomposing the process into manageable steps, showing that multi-dimensional limits can be tackled by examining one dimension at a time.

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus when dealing with indeterminate forms, like '0/0' or 'infinity/infinity'. To apply this rule, you must have a function that can be expressed as a fraction where the limits of both the numerator and the denominator approach zero or infinity. The rule states that under these conditions, the limit of the original function is the same as the limit of the derivatives of the numerator and denominator.

In the context of our exercise with the limit \( \lim_{r \rightarrow \infty} r \sin \frac{\pi}{r} \), we encounter an indeterminate form of 'infinity times 0'. Reconceiving this product as a quotient, we use L'Hopital's Rule, taking derivatives of the top and bottom until we obtain a determinate form. This method allowed us to successfully calculate the limit without getting stuck on the indeterminate form. Mastering L'Hopital's Rule gives students a robust method to handle limits that may not initially present a clear solution.

Indeterminate Forms

Indeterminate forms occur when evaluating a limit leads to an expression that is not immediately clear, such as '0/0', 'infinity/infinity', '0 times infinity', 'infinity minus infinity', '0 to the power of 0', 'infinity to the power 0', and '1 to the power infinity'. These forms are called indeterminate because they do not conclusively lead to a specific value without further analysis.

In real analysis, encountering an indeterminate form is a signal that additional techniques are needed to explore the limit's behavior. For instance, in our exercise, we navigated an indeterminate form by employing L'Hopital's Rule. Students should recognize these forms and know that they often require the application of limits laws, algebraic manipulation, or special rules like L'Hopital's to resolve the limit into a determinate value. Understanding indeterminate forms is crucial for students, as it equips them with the skills to handle complex limit problems that arise in higher mathematics.