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It follows that: lim n → → ∞ ∞ , n ∈ ∈ ( Z ) 1 ( 2 t ) 2 n + 1 = 1 + ∞ ∞ + 1 = 0 , | t | > 1 2 . {\displaystyle \lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\frac {1}{+\infty +1}}=0,|t|>{\tfrac {1}{2}}.}

Third, we consider the case where | t | = 1 2 . {\textstyle |t|={\frac {1}{2}}.} We may simply substitute in our equation:

lim n → → ∞ ∞ , n ∈ ∈ ( Z ) 1 ( 2 t ) 2 n + 1 = lim n → → ∞ ∞ , n ∈ ∈ ( Z ) 1 1 2 n + 1 = 1 1 + 1 = 1 2 . {\displaystyle \lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{1^{2n}+1}}={\frac {1}{1+1}}={\tfrac {1}{2}}.}

We see that it satisfies the definition of the pulse function. Therefore,

rect ⁡ ⁡ ( t ) = Π Π ( t ) = lim n → → ∞ ∞ , n ∈ ∈ ( Z ) 1 ( 2 t ) 2 n + 1 = { 0 if | t | > 1 2 1 2 if | t | = 1 2 1 if | t | < 1 2 . {\displaystyle \operatorname {rect} (t)=\Pi (t)=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1}}={\begin{cases}0&{\mbox{if }}|t|>{\frac {1}{2}}\\{\frac {1}{2}}&{\mbox{if }}|t|={\frac {1}{2}}\\1&{\mbox{if }}|t|<{\frac {1}{2}}.\\\end{cases}}}

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