已知等差数列\(\{a_{n}\}\)，公差\(d > 0\)，前\(n\)项和为\(S_{n}\)，且满足\(a_{2}a_{3}=45\)，\(a_{1}+a_{4}=14\)．
\((1)\)求数列\(\{a_{n}\}\)的通项公式及前\(n\)项和\(S_{n}\)；
\((2)\)设\(b_{n}= \dfrac {S_{n}}{n- \dfrac {1}{2}}\)，
\(①\)求证\(\{b_{n}\}\)是等差数列．
\(②\)求数列\(\{ \dfrac {1}{b_{n}\cdot b_{n+1}}\}\)的前\(n\)项和\(T_{n}\)．
\(③\)求\( \lim\limits_{n→∞}T_{n}\)．