2.8-bo'lim : Infinity chegaralari, II qism
2. \(f\left( x \right) = {{\bf{e}}^{\frac{{6{x^2} + x}}{{5 + 3x}}}}\) uchun quyidagi chegaralarning har birini baholang.
- \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\)
a \(\mathop {\lim }\limits_{x \to \, - \infty } f\left( x \right)\) Yechim
Avval e'tibor bering,
\[\mathop {\lim }\limits_{x \to \, - \infty } \frac{{6{x^2} + x}}{{5 + 3x}} = - \infty \]Agar siz ushbu cheklovga ishonchingiz komil bo'lmasa, avvalgi bo'limga qaytib, ushbu turdagi limitlarni amalga oshirishingiz mumkinligiga ishonch hosil qilish uchun u erdagi ba'zi misollar bilan ishlashingiz kerak.
Endi eslash 1-misol bu bo'limdan shuni bilamizki, ko'rsatkich chegarada manfiy cheksizlikka ketadi, chunki javob:
\[\mathop {\lim }\limits_{x \to \, - \infty } {{\bf{e}}^{\frac{{6{x^2} + x}}{{5 + 3x}}}} = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]b \(\mathop {\lim }\limits_{x \to \,\infty } f\left( x \right)\) Yechim
Avval e'tibor bering,
\[\mathop {\lim }\limits_{x \to \,\infty } \frac{{6{x^2} + x}}{{5 + 3x}} = \infty \]Agar siz ushbu cheklovga ishonchingiz komil bo'lmasa, avvalgi bo'limga qaytib, ushbu turdagi limitlarni amalga oshirishingiz mumkinligiga ishonch hosil qilish uchun u erdagi ba'zi misollar bilan ishlashingiz kerak.
Endi eslash 1-misol Ushbu bo'limdan biz bilamizki, ko'rsatkich chegarada cheksizlikka boradi, chunki javob:
\[\mathop {\lim }\limits_{x \to \,\infty } {{\bf{e}}^{\frac{{6{x^2} + x}}{{5 + 3x}}}} = \require{bbox} \bbox[2pt,border:1px solid black]{\infty }\]