Mana bu chegara uchun ish. Har bir bosqichda foydalanilgan xususiyat (yoki xususiyatlar) ro'yxatga olinadi va ba'zi hollarda xususiyatlar ko'rsatilgan bosqichda bir necha marta ishlatilgan bo'lishi mumkinligini unutmang.

\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to - 4} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{{h\left( x \right)}}{{f\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to - 4} \frac{{f\left( x \right)}}{{g\left( x \right)}} - \mathop {\lim }\limits_{x \to - 4} \frac{{h\left( x \right)}}{{f\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \frac{{\mathop {\lim }\limits_{x \to - 4} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to - 4} g\left( x \right)}} - \frac{{\mathop {\lim }\limits_{x \to - 4} h\left( x \right)}}{{\mathop {\lim }\limits_{x \to - 4} f\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 4}}\\ & = \frac{1}{{10}} - \frac{{ - 7}}{1} & & \hspace{0.25in}{\mbox{Plug in values of limits}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{71}}{{10}}}} & & & \end{alignat*}\]

E'tibor bering, ikkinchi bosqichda 4-xususiyatidan foydalanish imkoniga ega bo'ldik, chunki biz maxrajlar chegarasini (ikkalasi ham) baholaganimizdan so'ng, biz maxrajlarning chegaralari nolga teng emasligini aniqladik.

Mana bu chegara uchun ish. Har bir bosqichda foydalanilgan xususiyat (yoki xususiyatlar) ro'yxatga olinadi va ba'zi hollarda xususiyatlar ko'rsatilgan bosqichda bir necha marta ishlatilgan bo'lishi mumkinligini unutmang.

\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to - 4} \left[ {f\left( x \right)g\left( x \right)h\left( x \right)} \right] & = \left[ {\mathop {\lim }\limits_{x \to - 4} f\left( x \right)} \right]\left[ {\mathop {\lim }\limits_{x \to - 4} g\left( x \right)} \right]\left[ {\mathop {\lim }\limits_{x \to - 4} h\left( x \right)} \right] & & \hspace{0.25in}{\mbox{Property 3}}\\ & = \left( 1 \right)\left( {10} \right)\left( { - 7} \right) & & \hspace{0.25in}{\mbox{Plug in value of limits}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 70}} & & & \end{alignat*}\]

E'tibor bering, ushbu bo'limdagi 2 va 3-xususiyatlar faqat ikkita funktsiya bilan berilgan, ammo ular 3-xususiyat uchun qilganimizdek, osongina ikkitadan ko'proq funktsiyalarga kengaytirilishi mumkin.

Mana bu chegara uchun ish. Har bir bosqichda foydalanilgan xususiyat (yoki xususiyatlar) ro'yxatga olinadi va ba'zi hollarda xususiyatlar ko'rsatilgan bosqichda bir necha marta ishlatilgan bo'lishi mumkinligini unutmang.

\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to - 4} \left[ {\frac{1}{{h\left( x \right)}} + \frac{{3 - f\left( x \right)}}{{g\left( x \right) + h\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to - 4} \frac{1}{{h\left( x \right)}} + \mathop {\lim }\limits_{x \to - 4} \frac{{3 - f\left( x \right)}}{{g\left( x \right) + h\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \frac{{\mathop {\lim }\limits_{x \to - 4} 1}}{{\mathop {\lim }\limits_{x \to - 4} h\left( x \right)}} + \frac{{\mathop {\lim }\limits_{x \to - 4} \left[ {3 - f\left( x \right)} \right]}}{{\mathop {\lim }\limits_{x \to - 4} \left[ {g\left( x \right) + h\left( x \right)} \right]}} & & \hspace{0.25in}{\mbox{Property 4}}\\ & = \frac{{\mathop {\lim }\limits_{x \to - 4} 1}}{{\mathop {\lim }\limits_{x \to - 4} h\left( x \right)}} + \frac{{\mathop {\lim }\limits_{x \to - 4} 3 - \mathop {\lim }\limits_{x \to - 4} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to - 4} g\left( x \right) + \mathop {\lim }\limits_{x \to - 4} h\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \frac{1}{{ - 7}} + \frac{{3 - 1}}{{10 - 7}} & & \hspace{0.25in}{\mbox{Plug in values of limits }}\\ & & & \hspace{0.25in}{\mbox{\& Property 1}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{11}}{{21}}}} & & & \end{alignat*}\]

E'tibor bering, ikkinchi bosqichda 4-xususiyatidan foydalanish imkoniga ega bo'ldik, chunki biz maxrajlar chegarasini (ikkalasi ham) baholaganimizdan so'ng, biz maxrajlarning chegaralari nolga teng emasligini aniqladik.

Mana bu chegara uchun ish. Har bir bosqichda foydalanilgan xususiyat (yoki xususiyatlar) ro'yxatga olinadi va ba'zi hollarda xususiyatlar ko'rsatilgan bosqichda bir necha marta ishlatilgan bo'lishi mumkinligini unutmang.

\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to - 4} \left[ {2h\left( x \right) - \frac{1}{{h\left( x \right) + 7f\left( x \right)}}} \right] & = \mathop {\lim }\limits_{x \to - 4} 2h\left( x \right) - \mathop {\lim }\limits_{x \to - 4} \frac{1}{{h\left( x \right) + 7f\left( x \right)}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \mathop {\lim }\limits_{x \to - 4} 2h\left( x \right) - \frac{{\mathop {\lim }\limits_{x \to - 4} 1}}{{\mathop {\lim }\limits_{x \to - 4} \left[ {h\left( x \right) + 7f\left( x \right)} \right]}} & & \hspace{0.25in}{\mbox{Property 4}} & & & \end{alignat*}\]

Shu o'rinda bir daqiqa orqaga qadam qo'yaylik. Oldingi qismlarda biz ratsional ifodada 4-xususiyatdan foydalanish haqida tashvishlanmadik. Biroq, bu holatda, keling, biroz ehtiyotkor bo'laylik. Biz faqat maxraj chegarasi nolga teng bo'lmasa, 4-xususiyatdan foydalanishimiz mumkin. Keling, bu chegarani tekshirib ko'raylik va biz nimani olamiz.

\[\begin{alignat*}{3}\hspace{0.25in}\mathop {\lim }\limits_{x \to - 4} \left[ {h\left( x \right) + 7f\left( x \right)} \right] & = \mathop {\lim }\limits_{x \to - 4} h\left( x \right) + \mathop {\lim }\limits_{x \to - 4} \left[ {7f\left( x \right)} \right] & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \mathop {\lim }\limits_{x \to - 4} h\left( x \right) + 7\mathop {\lim }\limits_{x \to - 4} f\left( x \right) & & \hspace{0.25in}{\mbox{Property 1}}\\ & = - 7 + 7\left( 1 \right) & & \hspace{0.25in}{\mbox{Plug in values of limits \& Property 1}}\\ & = 0 & & & \end{alignat*}\]

Xo'sh, biz ikkinchi hadda maxraj chegarasi nolga teng bo'lishini ko'rishimiz mumkin, shuning uchun biz bu muddatda 4-xususiyatidan foydalana olmaymiz. Bu shuni anglatadiki, bu cheklovni amalga oshirish mumkin emas va shuni ta'kidlash kerakki, biz birinchi muddat chegarasi uchun qiymatni aniqlashimiz bu haqiqatni o'zgartirmaydi. Bu cheklovni amalga oshirish mumkin emas.