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The limit of a sequence of real numbers is $a$ if, for every $\epsilon >0$, there is an integer $N$ such that $n\ge N$ implies $|a_n-a|<\epsilon$. It tends to positive infinity if, for every real $M$, there is an integer $N$ such that $n\ge N$ implies $a_n>M$.

If $a_n\to +\infty$, choose the latter threshold with $M=1/\epsilon$. Then $a_n>1/\epsilon >0$ eventually, so $|1/a_n|<\epsilon$. Thus $1/a_n\to 0$. The converse is false: $a_n=-n$ has reciprocal tending to zero and tends to negative infinity. The precise sign-free statement is given by reciprocal limits and escape in absolute value: $1/a_n\to 0$ is equivalent to $|a_n|\to \infty$.

If $a_n\to a\ne 0$, eventually $|a_n-a|<|a|/2$, and the triangle inequality gives $|a_n|>|a|/2$. ConsequentlyGiven $\epsilon >0$, take $N$ large enough that $|a_n-a|<\min (|a|/2,\epsilon |a{|}^2/2)$ for all $n\ge N$. This proves $1/a_n\to 1/a$ directly from the definition of the limit of a sequence.