Hãy xác định xem kết quả nào dưới đây là sai ?
\(\lim\limits_{x\rightarrow3}\frac{x^2-4x+3}{x-3}=2\) \(\lim\limits_{x\rightarrow2}\dfrac{x^2+x-6}{x^2-4}=\dfrac{5}{4}\) \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(1-x\right)^2}=10\) \(\lim\limits_{x\rightarrow4}\frac{x^2-16}{x^2+x-20}=\frac{9}{8}\) Hướng dẫn giải:\(f\left(x\right)=\frac{x^2-4x+3}{x-3}=\frac{\left(x-3\right)\left(x-1\right)}{x-3}=x-1,\left(x\ne3\right)\);
\(g\left(x\right)=\frac{x^2+x-6}{x^2-4}=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x+3}{x+2},\left(\forall x\ne2\right)\);
\(h\left(x\right)=\frac{4x^6-5x^5+x}{\left(1-x\right)^2}=\frac{x\left(4x^5-5x^4+1\right)}{\left(x-1\right)^2}=\frac{x\left(x-1\right)\left(4x^4-x^3-x^2-x-1\right)}{\left(x-1\right)^2}=\frac{x\left(x-1\right)\left(4x^3+3x^2+2x+1\right)}{x-1}=x\left(4x^3+3x^2+2x+1\right),\left(\forall x\ne1\right)\)
\(k\left(x\right)=\frac{x^2-16}{x^2+x-20}=\frac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+5\right)}=\frac{x+4}{x+5}\left(x\ne4\right)\)
Nên \(\lim\limits_{x\rightarrow3}\frac{x^2-4x+3}{x-3}=\lim\limits_{x\rightarrow3}f\left(x\right)=\lim\limits_{x\rightarrow3}\left(x-1\right)=2;\) \(\lim\limits_{x\rightarrow2}\frac{x^2+x-6}{x^2-4}=\lim\limits_{x\rightarrow2}\frac{x+3}{x+2}=\frac{5}{4};\) \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(1-x\right)^2}=\lim\limits_{x\rightarrow1}k\left(x\right)=\lim\limits_{x\rightarrow1}x\left(4x^3+3x^2+2x+1\right)=10;\)
\(\lim\limits_{x\rightarrow4}\frac{x^2-16}{x^2+x-20}=\lim\limits_{x\rightarrow4}\frac{x+4}{x+5}=\frac{8}{9}\ne\frac{9}{8}\)
Kết quả sai là \(\lim\limits_{x\rightarrow4}\frac{x^2-16}{x^2+x-20}=\frac{9}{8}\)