Haqiqatan ham bu muammo juda ko'p emas. Ushbu bo'limda biz ko'rib chiqqan hisoblash limitlarining asosiy g'oyalarini eslang. Bunday holda, agar biz qiymatni kiritsak, biz 0/0 olishini ko'ramiz. Eslatib o'tamiz, bu limit mavjud emas degani EMAS. Xulosa qilishdan oldin yana bir oz ish qilishimiz kerak. Bu erda oddiy ko'paytuvchilarga ajratish bizga unchalik foyda keltirmaydi, shuning uchun bu holda biz suratni ratsionalizatsiya qilishimiz kerak bo'ladi.

\[\begin{align*}\mathop {\lim }\limits_{x \to \, - 3} \frac{{\sqrt {2x + 22} - 4}}{{x + 3}} & = \mathop {\lim }\limits_{x \to \, - 3} \frac{{\left( {\sqrt {2x + 22} - 4} \right)}}{{\left( {x + 3} \right)}}\frac{{\left( {\sqrt {2x + 22} + 4} \right)}}{{\left( {\sqrt {2x + 22} + 4} \right)}} = \mathop {\lim }\limits_{x \to \, - 3} \frac{{2x + 22 - 16}}{{\left( {x + 3} \right)\left( {\sqrt {2x + 22} + 4} \right)}}\\ & = \mathop {\lim }\limits_{x \to \, - 3} \frac{{2\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {\sqrt {2x + 22} + 4} \right)}} = \mathop {\lim }\limits_{x \to \, - 3} \frac{2}{{\sqrt {2x + 22} + 4}} = \frac{2}{8} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{4}}}\end{align*}\]