Haqiqatan ham bu muammo bilan bu erda ko'p narsa qilish mumkin emas. Biz qilishimiz kerak bo'lgan narsa vektorning barcha komponentlarining chegarasini olishdir.

\[\begin{align*}\mathop {\lim }\limits_{t \to 1} \left\langle {{{\bf{e}}^{t - 1}},4t,\frac{{t - 1}}{{{t^2} - 1}}} \right\rangle & = \left\langle {\mathop {\lim }\limits_{t \to 1} {{\bf{e}}^{t - 1}},\mathop {\lim }\limits_{t \to 1} 4t,\mathop {\lim }\limits_{t \to 1} \frac{{t - 1}}{{{t^2} - 1}}} \right\rangle \\ & = \left\langle {\mathop {\lim }\limits_{t \to 1} {{\bf{e}}^{t - 1}},\mathop {\lim }\limits_{t \to 1} 4t,\mathop {\lim }\limits_{t \to 1} \frac{1}{{2t}}} \right\rangle = \left\langle {{{\bf{e}}^0},4,\frac{1}{2}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {1,4,\frac{1}{2}} \right\rangle }}\end{align*}\]

L'Hospital qoidasini unutmang! Bizga uchinchi muddat uchun kerak edi.