Assume [tex]\dfrac1{a_n}[/tex] is not bounded, i.e. there are no [tex]u,\ell[/tex] for which [tex]\ell\le\dfrac1{a_n}\le u[/tex] for all [tex]n[/tex].
Now, [tex]a_n\to k\neq0[/tex] is to say that for any [tex]\varepsilon>0[/tex], we can find a large enough [tex]N[/tex] such that [tex]|a_n-k|<\varepsilon[/tex] whenever [tex]n\ge N[/tex]. Simultaneously, this means that [tex]a_n[/tex] is bounded.
Let's suppose without loss of generality that [tex]a_n\neq0[/tex] for any [tex]n[/tex]. (Note that if [tex]a_n=0[/tex] for some finite number of values of [tex]n[/tex], we can simply remove them from consideration.)
So we have
[tex]|a_n-k|=\left|a_n\left(1-\dfrac k{a_n}\right)\right|=|a_n|\left|1-\dfrac k{a_n}\right|<\varepsilon[/tex]
Because [tex]a_n[/tex] is bounded, we know there is some [tex]\alpha[/tex] such that [tex]|a_n|\le\alpha[/tex] for all [tex]n[/tex]. Now,
[tex]|a_n|\left|1-\dfrac k{a_n}\right|\le\alpha\left|1-\dfrac k{a_n}\right|<\varepsilon[/tex]
[tex]\left|1-\dfrac k{a_n}\right|<\dfrac\varepsilon\alpha[/tex]
[tex]-\dfrac\varepsilon\alpha<1-\dfrac k{a_n}<\dfrac\varepsilon\alpha[/tex]
[tex]-1-\dfrac\varepsilon\alpha<-\dfrac k{a_n}<-1+\dfrac\varepsilon\alpha[/tex]
[tex]1-\dfrac\varepsilon\alpha<\dfrac k{a_n}<1+\dfrac\varepsilon\alpha[/tex]
[tex]\dfrac1k-\dfrac\varepsilon{\alpha k}<\dfrac 1{a_n}<\dfrac1k+\dfrac\varepsilon{\alpha k}[/tex]
But we initially assumed that [tex]\dfrac1{a_n}[/tex] is unbounded, so the above is impossible. Thus [tex]\dfrac1{a_n}[/tex] must be bounded.
