If [tex](a_n) \rightarrow 0[/tex] and [tex]\left|b_n - b\right| \leq a_n[/tex], then [tex](b_n) \rightarrow b[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

I'm a little bit stuck on proving this! Part (iii) of the Order Limit Theorem states:

Assume:[tex]lim a_n = a[/tex] and [tex]lim b_n = b[/tex]

If there exists [tex]c \in[/tex]Rfor which [tex]c \leq b_n[/tex] for all [tex]n \in[/tex]N, then [tex]c \leq b[/tex]. Similarly if [tex]a_n \leq c[/tex] for all [tex]n \in[/tex]N, then [tex]a \leq c[/tex].

I can then take this to a point where [tex]b_n = b[/tex], but somehow that doesn't seem right because [tex]\left|b_n - b\right|<\epsilon[/tex], where [tex]\epsilon > 0[/tex]. Where am I going wrong here!

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# Proving a limit and the Order Limit Theorem

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