= Solution
The <limit of a sequence> of <real numbers> is $a$ if, for every $\varepsilon>0$, there is an <integer> $N$ such that $n\ge N$ implies $|a_n-a|<\varepsilon$. 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 $M=1/\varepsilon$. Then $a_n>1/\varepsilon>0$ eventually, so $|1/a_n|<\varepsilon$. Thus $1/a_n\to0$. \b[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>: $1/a_n\to0$ is equivalent to $|a_n|\to\infty$.
If $a_n\to a\ne0$, eventually $|a_n-a|<|a|/2$, and the <triangle inequality> gives $|a_n|>|a|/2$. Consequently
$$
\left|\frac1{a_n}-\frac1a\right|
=\frac{|a_n-a|}{|a_n||a|}\le\frac{2|a_n-a|}{|a|^2}.
$$
Given $\varepsilon>0$, take $N$ large enough that $|a_n-a|<\min(|a|/2,\varepsilon|a|^2/2)$ for all $n\ge N$. This proves \b[$1/a_n\to1/a$] directly from the definition of the <limit of a sequence>.
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