Answer to Question #343661 in Real Analysis for Mr Alex

Question #343661

a) Does the sequence (3+(-1)n) converge to 2? Justify.

 

 b) Show that "\\lim _{x\\to \\infty }\\left(\\frac{x-3}{x+1}\\right)^x=e^{-4}"


c) Check whether the sequence fn (x) = "\\frac{3x}{1+nx^2}" where x ∈ [2,∞ [ is uniformly 

convergent in [2,∞ [


1
Expert's answer
2022-05-25T15:21:32-0400

1) The sequence 

"\\left(3+\\left(-1\\right)^n\\right)"

doesn't converge to 2, because this sequence is oscillating between 2 and 4. Hence, it is divergent.


2) "\\lim_{x\\rightarrow\\infty}{\\left(\\frac{x-3}{x+1}\\right)^x}=\\lim_{x\\rightarrow\\infty}{\\left(\\frac{x+1-4}{x+1}\\right)^x}=" "\\lim\\limits_{x\\rightarrow\\infty}{\\left(\\left(1+\\frac{-4}{x+1}\\right)^\\frac{x+1}{-4}\\right)^{-\\frac{4x}{x+1}}}="

"=\\lim_{x\\rightarrow\\infty}{e^{-\\frac{4}{1+\\frac{1}{x}}}}= e^{-4}".


3) if x=2 then "f_n\\left(2\\right)=\\frac{6}{1+4n}" and "\\lim_{n\\rightarrow\\infty}{f_n\\left(2\\right)}=\\lim_{n\\rightarrow\\infty}{\\frac{6}{1+4n}}=0".


If x>2 then "\\left|f_n\\left(x\\right)\\right|=\\left|\\frac{3x}{1+nx^2}\\right|\\le\\left|\\frac{3x}{nx^2}\\right|=\\left|\\frac{3}{nx}\\right|" and "\\lim_{n\\rightarrow\\infty}{\\frac{3}{nx}}=0"

Conclution:

"f_n\\left(x\\right)\\rightrightarrows0,\\ x\\in\\left[2,+\\infty\\right["

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