Prove that the limit of the functions

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SUMMARY

The limit of the functions cos(x), sin(x), and e^{ix} as x approaches infinity is 0. The discussion clarifies that e^{ix} is bounded, with both its real and imaginary components not exceeding 1. Additionally, the limit of the expression (1+x)^i as x tends to infinity is also addressed, confirming that e^{(i-1)x} approaches zero due to the e^{-x} component dominating the behavior of the function.

PREREQUISITES
  • Understanding of trigonometric functions: cos(x) and sin(x)
  • Familiarity with complex exponentials: e^{ix}
  • Knowledge of limits in calculus
  • Basic understanding of complex numbers and their properties
NEXT STEPS
  • Study the properties of bounded functions in complex analysis
  • Learn about limits involving complex exponentials and their implications
  • Explore the behavior of functions as x approaches infinity in calculus
  • Investigate the implications of Euler's formula: e^{ix} = cos(x) + i*sin(x)
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Students of calculus, mathematicians, and anyone interested in complex analysis or the behavior of trigonometric functions at infinity.

eljose
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how do you prove that the limit of the functions [tex]cos(x) ,sen(x), e^{ix}[/tex] is 0 when [tex]x\rightarrow\infty[/tex]

another question what would be the limit of [tex](1+x)^i[/tex] tending x to infinite?..thanx
 
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I don't know what sen is but cos(x) and eix have no limit as x approches infinity.

Same goes for your other problem as well.
 
Note that
[tex]e^{ix}[/tex]
is bounded:
[tex]e^{ix} = \cos(x) + i\sin(x)[/tex]
so, for large, or arbitrary x, e^{ix}, the imaginary part will never be greater than 1, and the real part will never be greater than 1.
Thus, if you had:
[tex]\lim_{x \rightarrow \infty} e^{(i-1)x}[/tex]
the e^{-x} part forces the whole thing to go to zero.
 

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