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$\underset{x\to \mathrm{\infty }}{lim}\left(\frac{4}{5+\mathrm{cos}x}\right)$ prove that the limit is not finite while $x\to \mathrm{\infty }$ using Cauchy definition
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stacan6t
Suppose that limit L exists. For $ϵ=\frac{1}{10}$ such that $x>{M}_{1}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(x\right)-L|<\frac{1}{10}$
Let ${N}_{1}\in \mathbb{N}$ be such that $2\pi {N}_{1}>{M}_{1}$. Then if
$n>{N}_{1}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}2n\pi >2\pi {N}_{1}>{M}_{1}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(2n\pi \right)-L|<\frac{1}{10}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|\frac{2}{3}-L|=|\frac{4}{5+\mathrm{cos}\left(2n\pi \right)}-L|<\frac{1}{10}$ Next for
$ϵ=\frac{1}{15},\mathrm{\exists }{M}_{2}$ such that $x>{M}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(x\right)-L|<\frac{1}{15}$. Again, let ${N}_{2}\in \mathbb{N}$ be such that $\left(2{N}_{2}+1\right)\pi >{M}_{2}$. So if
$n>{N}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(2n+1\right)\pi >\left(2{N}_{2}+1\right)\pi >{M}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(\left(2n+1\right)\pi \right)-L|<\frac{1}{15}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|1-L|<\frac{1}{15}$
Thus if $n>{N}_{1}+{N}_{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3}\le |1-L|+|\frac{2}{3}-L|<\frac{1}{10}+\frac{1}{15}=\frac{1}{6}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3}<\frac{1}{6}$
Contradiction. So the limit Lcannot exists