Tìm phương trình đường tiệm cận ngang của đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaG4maiaadIhacqGHRaWkcaaIYaaabaGaamiEaiab % gUcaRiaaigdaaaaaaa!3DFA! y = \frac{{3x + 2}}{{x + 1}}\).
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Lời giải:
Báo saiTa có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaca % qGSbGaaeyAaiaab2gaaSqaaiaadIhacqGHsgIRcqGHRaWkcqGHEisP % aeqaaOWaaSaaaeaacaaIZaGaamiEaiabgUcaRiaaikdaaeaacaWG4b % Gaey4kaSIaaGymaaaacqGH9aqpaaa!4547! \mathop {{\rm{lim}}}\limits_{x \to + \infty } \frac{{3x + 2}}{{x + 1}} = \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaca % qGSbGaaeyAaiaab2gaaSqaaiaadIhacqGHsgIRcqGHRaWkcqGHEisP % aeqaaOWaaSaaaeaacaaIZaGaey4kaSYaaSaaaeaacaaIYaaabaGaam % iEaaaaaeaacaaIXaGaey4kaSYaaSaaaeaacaaIXaaabaGaamiEaaaa % aaGaeyypa0JaaG4maaaa!46DF! \mathop {{\rm{lim}}}\limits_{x \to + \infty } \frac{{3 + \frac{2}{x}}}{{1 + \frac{1}{x}}} = 3\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yEaiabg2da9iaaiodaaaa!3B12! \Rightarrow y = 3\) là tiệm cận nganh của đồ thị hàm số.