2.6-bo'lim : Cheksiz limitlar
8. \(\displaystyle g\left( x \right) = \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}\) ning barcha vertikal asimptotalarini toping.
Yechim
Eslatib o'tamiz, \(x = a\) da vertikal asimptotlar paydo bo'ladi, agar \(x = a\) dagi chegaralardan birortasi (bir tomonlama yoki umumiy chegara) ortiqcha yoki minus cheksizlik bo'lsa.
Oldingi misollardan ko'rishimiz mumkinki, ratsional ifodalar uchun nolga bo'linish mavjud bo'lganda vertikal asimptotlar paydo bo'ladi. Shunday qilib, biz \(x = - 5\) va \(x = 9\) da mumkin bo'lgan vertikal asimptotaga ega bo'lamiz.
Keling, ularning har biri uchun ikkita bir tomonlama chegaralarni baholab, ularning aslida vertikal asimptotlar ekanligini tekshirib ko'raylik.
\(x = - 5\) dan boshlaylik. Biz baholashimiz kerak,
\[\mathop {\lim }\limits_{x \to \, - {5^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}\hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \, - {5^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}}\]Ikkala holatda ham \(x \to - 5\) (chapdan ham, o'ngdan ham) surat doimiy -8 bo'ladi.
Bir tomonlama limitlar uchun bizda quyidagi ma'lumotlar mavjud:
\[\begin{align*}x \to - {5^ - }\hspace{0.25in} \Rightarrow \hspace{0.25in}x < - 5\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x + 5 < 0\\ x \to - {5^ + }\hspace{0.25in} \Rightarrow \hspace{0.25in}x > - 5\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x + 5 > 0\end{align*}\]Shuni ham yodda tutingki, \(x\) -5 ga etarlicha yaqin bo'lsa (biz \(x \to - 5\) ni ko'rib chiqayotganimiz uchun xavfsiz deb hisoblashimiz mumkin), bizda \(x - 9 < 0\) bo'ladi.
Shunday qilib, chap chegarada surat qat'iy manfiy son, maxraj esa ijobiy (ikki manfiy sonning ko'paytmasi) va tobora kichikdir. Xuddi shunday, o'ng chegara uchun maxraj manfiy (musbat va manfiy sonning mahsuloti) va tobora kichikroq. Shunday qilib, biz ega bo'lamiz,
\[\mathop {\lim }\limits_{x \to \, - {5^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = - \infty \hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \, - {5^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = \infty \]Endi \(x = 9\) uchun. Shunga qaramay, surat doimiy -8 dir. Bizda ham bor,
\[\begin{align*}x \to {9^ - }\hspace{0.25in} \Rightarrow \hspace{0.25in}x < 9\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x - 9 < 0\\ x \to {9^ + }\hspace{0.25in} \Rightarrow \hspace{0.25in}x > 9\hspace{0.25in}\, \Rightarrow \hspace{0.25in}x - 9 > 0\end{align*}\]Nihoyat, \(x\) 9 ga etarlicha yaqin bo'lsa (biz \(x \to 9\) ni ko'rib chiqayotganimiz uchun xavfsiz deb hisoblashimiz mumkin), bizda \(x + 5 > 0\) bo'ladi.
Shunday qilib, chap chegarada surat qat'iy manfiy son va maxraj manfiy (musbat va manfiy sonning ko'paytmasi) va borgan sari kichikdir. Xuddi shunday, o'ng chegara uchun maxraj musbat (ikkita ijobiy sonning ko'paytmasi) va tobora kichikroq. Shunday qilib, biz ega bo'lamiz,
\[\mathop {\lim }\limits_{x \to \,{9^{\, - }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = \infty \hspace{0.25in}\,\,\,\,{\rm{and}}\hspace{0.25in}\hspace{0.25in}\mathop {\lim }\limits_{x \to \,{9^{\, + }}} \frac{{ - 8}}{{\left( {x + 5} \right)\left( {x - 9} \right)}} = - \infty \]Shunday qilib, ushbu limitlarning barchasi ko'rsatganidek, bizda \(x = - 5\) va \(x = 9\) da vertikal asimptotlar mavjud.