lim,... sin(7) does not exist. 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Use the definition of Convergent Sequences to prove:
![The expression depicted is a mathematical limit problem:
\[
\lim_{{n \to \infty}} \sin\left(\frac{n\pi}{2}\right) \text{ does not exist.}
\]
### Explanation:
This statement involves the concept of limits in trigonometry and sine functions. To understand why the limit does not exist, consider:
1. **Inside Sine Function**: The term \(\frac{n\pi}{2}\) generates angles in radians that are multiples of \(\pi\). As \(n\) takes integer values such as 0, 1, 2, 3, etc., the resulting angles will be 0, \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), \(2\pi\), etc.
2. **Values of Sine**: The sine of these angles follows a pattern:
- \(\sin(0) = 0\)
- \(\sin\left(\frac{\pi}{2}\right) = 1\)
- \(\sin(\pi) = 0\)
- \(\sin\left(\frac{3\pi}{2}\right) = -1\)
- \(\sin(2\pi) = 0\)
3. **Oscillation**: The output of the sine function oscillates between -1, 0, and 1 continually as \(n\) increases. Thus, the sequence of sine values does not approach a single value as \(n\) approaches infinity.
### Conclusion:
Since the sequence does not settle at a particular value, the limit of \(\sin\left(\frac{n\pi}{2}\right)\) as \(n\) approaches infinity does not exist. This illustrates the behavior of trigonometric functions when approached through limits involving infinity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe00cebfb-aec4-474d-8979-79a2d105b819%2Fd6414923-ed68-47b8-a8e0-cf2c6fece0d3%2Fe5o4ye_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The expression depicted is a mathematical limit problem:
\[
\lim_{{n \to \infty}} \sin\left(\frac{n\pi}{2}\right) \text{ does not exist.}
\]
### Explanation:
This statement involves the concept of limits in trigonometry and sine functions. To understand why the limit does not exist, consider:
1. **Inside Sine Function**: The term \(\frac{n\pi}{2}\) generates angles in radians that are multiples of \(\pi\). As \(n\) takes integer values such as 0, 1, 2, 3, etc., the resulting angles will be 0, \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), \(2\pi\), etc.
2. **Values of Sine**: The sine of these angles follows a pattern:
- \(\sin(0) = 0\)
- \(\sin\left(\frac{\pi}{2}\right) = 1\)
- \(\sin(\pi) = 0\)
- \(\sin\left(\frac{3\pi}{2}\right) = -1\)
- \(\sin(2\pi) = 0\)
3. **Oscillation**: The output of the sine function oscillates between -1, 0, and 1 continually as \(n\) increases. Thus, the sequence of sine values does not approach a single value as \(n\) approaches infinity.
### Conclusion:
Since the sequence does not settle at a particular value, the limit of \(\sin\left(\frac{n\pi}{2}\right)\) as \(n\) approaches infinity does not exist. This illustrates the behavior of trigonometric functions when approached through limits involving infinity.
Transcribed Image Text:The text in the image reads:
"The limit as n approaches infinity of 3 times (-1) raised to the power of n does not exist."
Explanation:
The expression involves an alternating sequence determined by the term \((-1)^n\), which switches between -1 and 1 as \(n\) changes from even to odd numbers. As \(n\) approaches infinity, the sequence 3(-1)^n oscillates between 3 and -3. Due to this oscillation, the limit does not converge to a single value, and therefore, it does not exist.
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