2.4-bo'lim : Limit xususiyatlari
3. Berilgan \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 6\), \(\mathop {\lim }\limits_{x \to 0} g\left( x \right) = - 4\) va \(\mathop {\lim }\limits_{x \to 0} h\left( x \right) = - 1\) quyidagi chegaralarning har birini hisoblash uchun ushbu bo'limda berilgan limit xususiyatlaridan foydalanadi. Agar biron bir chegarani hisoblashning iloji bo'lmasa, nima uchun buni aniq tushuntiring.
- \(\mathop {\lim }\limits_{x \to 0} {\left[ {f\left( x \right) + h\left( x \right)} \right]^3}\)
- \(\mathop {\lim }\limits_{x \to 0} \sqrt {g\left( x \right)h\left( x \right)} \)
- \(\mathop {\lim }\limits_{x \to 0} \sqrt[3]{{11 + {{\left[ {g\left( x \right)} \right]}^2}}}\)
- \(\displaystyle \mathop {\lim }\limits_{x \to 0} \sqrt {\frac{{f\left( x \right)}}{{h\left( x \right) - g\left( x \right)}}} \)
a \(\mathop {\lim }\limits_{x \to 0} {\left[ {f\left( x \right) + h\left( x \right)} \right]^3}\) Yechim
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.
\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to 0} {\left[ {f\left( x \right) + h\left( x \right)} \right]^3} & = {\left[ {\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right) + h\left( x \right)} \right)} \right]^3} & & \hspace{0.25in}{\mbox{Property 5}}\\ & = {\left[ {\mathop {\lim }\limits_{x \to 0} f\left( x \right) + \mathop {\lim }\limits_{x \to 0} h\left( x \right)} \right]^3} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = {\left[ {6 - 1} \right]^3} & & \hspace{0.25in}{\mbox{Plug in values of limits}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{125}} & & & \end{alignat*}\]b \(\mathop {\lim }\limits_{x \to 0} \sqrt {g\left( x \right)h\left( x \right)} \) Yechim
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.
\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to 0} \sqrt {g\left( x \right)h\left( x \right)} & = \sqrt {\mathop {\lim }\limits_{x \to 0} g\left( x \right)h\left( x \right)} & & \hspace{0.25in}{\mbox{Property 6}}\\ & = \sqrt {\left[ {\mathop {\lim }\limits_{x \to 0} g\left( x \right)} \right]\left[ {\mathop {\lim }\limits_{x \to 0} h\left( x \right)} \right]} & & \hspace{0.25in}{\mbox{Property 3}}\\ & = \sqrt {\left( { - 4} \right)\left( { - 1} \right)} & & \hspace{0.25in}{\mbox{Plug in value of limits}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{2} & & & \end{alignat*}\]c \(\mathop {\lim }\limits_{x \to 0} \sqrt[3]{{11 + {{\left[ {g\left( x \right)} \right]}^2}}}\) Yechim
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.
\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to 0} \sqrt[3]{{11 + {{\left[ {g\left( x \right)} \right]}^2}}} & = \sqrt[3]{{\mathop {\lim }\limits_{x \to 0} \left( {11 + {{\left[ {g\left( x \right)} \right]}^2}} \right)}} & & \hspace{0.25in}{\mbox{Property 6}}\\ & = \sqrt[3]{{\mathop {\lim }\limits_{x \to 0} 11 + \mathop {\lim }\limits_{x \to 0} {{\left[ {g\left( x \right)} \right]}^2}}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \sqrt[3]{{\mathop {\lim }\limits_{x \to 0} 11 + {{\left[ {\mathop {\lim }\limits_{x \to 0} g\left( x \right)} \right]}^2}}} & & \hspace{0.25in}{\mbox{Property 5}}\\ & = \sqrt[3]{{11 + {{\left( { - 4} \right)}^2}}} & & \hspace{0.25in}{\mbox{Plug in values of limits \& Property 7}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{3} & & & \end{alignat*}\]d \(\displaystyle \mathop {\lim }\limits_{x \to 0} \sqrt {\frac{{f\left( x \right)}}{{h\left( x \right) - g\left( x \right)}}} \) Yechim
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.
\[\begin{alignat*}{3}\mathop {\lim }\limits_{x \to 0} \sqrt {\frac{{f\left( x \right)}}{{h\left( x \right) - g\left( x \right)}}} & = \sqrt {\mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right)}}{{h\left( x \right) - g\left( x \right)}}} & & \hspace{0.25in}{\mbox{Property 6}}\\ & = \sqrt {\frac{{\mathop {\lim }\limits_{x \to 0} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {h\left( x \right) - g\left( x \right)} \right)}}} & & \hspace{0.25in}{\mbox{Property 4}}\\ & = \sqrt {\frac{{\mathop {\lim }\limits_{x \to 0} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 0} h\left( x \right) - \mathop {\lim }\limits_{x \to 0} g\left( x \right)}}} & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \sqrt {\frac{6}{{ - 1 - \left( { - 4} \right)}}} & & \hspace{0.25in}{\mbox{Plug in values of limits }}\\ & = \sqrt 2 & & & \end{alignat*}\]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.