Giới hạn hàm số

H

huynhbachkhoa23

lim⁡x→2+x2−9x+20x3−6x2+11x−6=lim⁡x→2+(x−4)(x−5)(x−1)(x−2)(x−3)=−∞\lim\limits_{x\to 2^{+}} \dfrac{x^2-9x+20}{x^3-6x^2+11x-6}=\lim\limits_{x\to 2^{+}} \dfrac{(x-4)(x-5)}{(x-1)(x-2)(x-3)}=-∞ do x−2>0x-2>0 và lim⁡x→2+(x−4)(x−5)(x−1)(x−3)=−6<0\lim\limits_{x\to 2^{+}}\dfrac{(x-4)(x-5)}{(x-1)(x-3)} =-6<0
lim⁡x→2−x2−9x+20x3−6x2+11x−6=lim⁡x→2−(x−4)(x−5)(x−1)(x−2)(x−3)=+∞\lim\limits_{x\to 2^{-}} \dfrac{x^2-9x+20}{x^3-6x^2+11x-6}=\lim\limits_{x\to 2^{-}} \dfrac{(x-4)(x-5)}{(x-1)(x-2)(x-3)}=+∞ do x−2<0x-2<0 và lim⁡x→2+(x−4)(x−5)(x−1)(x−3)=−6<0\lim\limits_{x\to 2^{+}}\dfrac{(x-4)(x-5)}{(x-1)(x-3)} =-6<0
Vậy không tồn tại giới hạn lim⁡x→2x2−9x+20x2−6x2+11x−6\lim\limits_{x\to 2}\dfrac{x^2-9x+20}{x^2-6x^2+11x-6}
 
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