Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Hello there, can you help me solve problems? Thanks!

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.
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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.
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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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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