\nA sucess\u00e3o de termo geral ${u_n} = \\operatorname{sen} \\left( {\\frac{{n\\pi }}{2}} \\right)$ \u00e9 limitada? E mon\u00f3tona?<\/p>\n<\/blockquote>\n
\u00ad<\/p>\n
A sucess\u00e3o $\\left( {{u_n}} \\right)$ \u00e9 a restri\u00e7\u00e3o a $\\mathbb{N}$ da fun\u00e7\u00e3o \\[\\begin{array}{*{20}{l}}
\n{f:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\operatorname{sen} \\left( {\\frac{\\pi }{2}x} \\right)}
\n\\end{array}\\]<\/p>\n
que possui contradom\u00ednio $D{‘_f} = \\left[ { – 1,1} \\right]$.<\/p>\n
\u00ad\u00ad
\nSeja $S$ o conjunto dos termos de $\\left( {{u_n}} \\right)$.<\/p>\n
Como a sucess\u00e3o $\\left( {{u_n}} \\right)$ \u00e9 a restri\u00e7\u00e3o de $f$ a $\\mathbb{N}$, ent\u00e3o $S \\subset \\left[ { – 1,1} \\right]$.<\/p>\n
Logo, a sucess\u00e3o $\\left( {{u_n}} \\right)$ \u00e9 limitada, pois \u00e9 limitado o conjunto dos seus termos.<\/p>\n
Note que $S = \\left\\{ { – 1,0,1} \\right\\}$.<\/p>\n
\u00ad\u00ad
\nA sucess\u00e3o $\\left( {{u_n}} \\right)$ n\u00e3o \u00e9 mon\u00f3tona, pois ${u_1} > {u_2} > {u_3} < {u_4}$, visto que ${u_1} = 1$, ${u_2} = 0$, ${u_3} =\u00a0 – 1$ e ${u_4} = 0$.<\/p>\n
\u00ad\u00ad
\n