Problem 12 - Thomas Calculus

Question

Find the limits in Exercises \(9-12\) . $$\lim _{r \rightarrow \infty} \frac{r+\sin r}{2 r+7-5 \sin r}$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Identify the Limit Expression

We're given the limit to find: \( \lim _{r \rightarrow \infty} \frac{r+\sin r}{2r+7-5\sin r} \). We need to analyze the behavior of this expression as \( r \) approaches infinity.

Step 2: Separate Dominant Terms

In the expression, the dominant terms are \( r \) in the numerator and \( 2r \) in the denominator because \( \sin r \) oscillates between -1 and 1 and becomes negligible at infinity.

Step 3: Divide by the Highest Power of \( r \)

To simplify the fraction \( \frac{r+\sin r}{2r+7-5\sin r} \), divide both numerator and denominator by \( r \), the highest power in the terms:\[\frac{r+\sin r}{r} = 1 + \frac{\sin r}{r},\]\[\frac{2r+7-5\sin r}{r} = 2 + \frac{7}{r} - \frac{5\sin r}{r}.\]

Step 4: Evaluate the Limit

As \( r \to \infty \), the terms \( \frac{\sin r}{r}, \frac{7}{r}, \) and \( \frac{5\sin r}{r} \) all approach 0. Thus, the limit of the expression simplifies to:\[\lim _{r \to \infty} \frac{1 + \frac{\sin r}{r}}{2 + \frac{7}{r} - \frac{5\sin r}{r}} = \frac{1+0}{2+0-0} = \frac{1}{2}.\]

Step 5: Conclude the Result

Thus, the limit evaluates to \( \frac{1}{2} \). We have shown how each part contributes to this final value.

Key Concepts

Dominant Terms

When we talk about dominant terms in a linear expression, these are the parts of the equation that grow the fastest as variables reach higher values such as infinity. In the provided limit problem, the expression is \( \lim _{r \rightarrow \infty} \frac{r+\sin r}{2r+7-5\sin r} \). Here, identifying the dominant terms means recognizing which parts of the expression have the greatest influence at extreme values of \( r \).

In this case:

• The term \( r \) in the numerator is dominant over \( \sin r \) because \( \sin r \) oscillates between -1 and 1, making it relatively inconsequential as \( r \) becomes very large.
• In the denominator, \( 2r \) outweighs both \( 7 \) and the oscillating \( 5\sin r \) for the same reason. The constant term \( 7 \) and any value multiplied with \( \sin r \) become negligible at infinity.

Identifying the dominant terms allows us to see where the majority of the 'weight' of the expression lies, enabling simplification by isolating these terms and considering their limits.

Infinity

Infinity, in the context of calculus and limits, represents an abstract concept where variables grow without bound. When we evaluate limits as a variable approaches infinity, as in our exercise, we're investigating how different functions behave as they grow very large or very small.

In the expression \( \lim_{r \rightarrow \infty} \frac{r + \sin r}{2r + 7 - 5 \sin r} \), analyzing it involves understanding how each component behaves as \( r \) becomes extremely large. Essentially:

• Terms like \( \frac{\sin r}{r} \) become insignificant because the denominator grows far faster than the oscillating function, forcing the fraction towards zero.
• Constant values such as \( 7 \) similarly become less significant compared with terms involving \( r \) because they do not change with \( r \).

As we substitute and analyze these behaviors with infinity, we observe that simplification often leads to clearer, more manageable comparisons when dominant terms are the focus, assisting in understanding the behavior of the whole function.

Oscillating Functions

Oscillating functions are those that repeatedly move between a set of upper and lower values. A classic example of an oscillating function is \( \sin r \), which fluctuates between -1 and 1 indefinitely. This property has a significant impact when finding limits, especially in terms of behavior as variables approach infinity or zero.

Let's consider the exercise's expression: \( \frac{r+\sin r}{2 r+7-5 \sin r} \). Since \( \sin r \) oscillates, it does not grow in magnitude as \( r \) increases. This causes:

• Minimal impact on the dominant term comparison as \( r \to \infty \), so \( \sin r \) becomes irrelevant in the limit since its effect diminishes due to its bounded nature.
• Even if multiplied by coefficients (like \(-5\sin r\) in the denominator), the bounded nature holds, shrugging off their influence in comparison to other growing terms like \( r \).

Understanding oscillating functions helps us comprehend why certain terms can be disregarded in limits, simplifying our path to understanding complex expressions.