find the following infinite limit. $lim_{x\rightarrowinfty}(\frac{(x - 2)^2(x - 3)}{(x^2 - 4)(x^2 - 6x + 9)})$

Explanation:

Step1: Expand the numerator and denominator

The numerator ((x - 2)^2(x - 3)=(x^{2}-4x + 4)(x - 3)=x^{3}-3x^{2}-4x^{2}+12x + 4x-12=x^{3}-7x^{2}+16x - 12). The denominator ((x^{2}-4)(x^{2}-6x + 9)=x^{4}-6x^{3}+9x^{2}-4x^{2}+24x - 36=x^{4}-6x^{3}+5x^{2}+24x - 36).

Step2: Divide both numerator and denominator by (x^{4})

(\lim_{x\rightarrow\infty}\frac{x^{3}-7x^{2}+16x - 12}{x^{4}-6x^{3}+5x^{2}+24x - 36}=\lim_{x\rightarrow\infty}\frac{\frac{x^{3}}{x^{4}}-\frac{7x^{2}}{x^{4}}+\frac{16x}{x^{4}}-\frac{12}{x^{4}}}{\frac{x^{4}}{x^{4}}-\frac{6x^{3}}{x^{4}}+\frac{5x^{2}}{x^{4}}+\frac{24x}{x^{4}}-\frac{36}{x^{4}}}) (=\lim_{x\rightarrow\infty}\frac{\frac{1}{x}-\frac{7}{x^{2}}+\frac{16}{x^{3}}-\frac{12}{x^{4}}}{1-\frac{6}{x}+\frac{5}{x^{2}}+\frac{24}{x^{3}}-\frac{36}{x^{4}}})

Step3: Evaluate the limit

As (x\rightarrow\infty), (\frac{1}{x}\rightarrow0), (\frac{7}{x^{2}}\rightarrow0), (\frac{16}{x^{3}}\rightarrow0), (\frac{12}{x^{4}}\rightarrow0), (\frac{6}{x}\rightarrow0), (\frac{5}{x^{2}}\rightarrow0), (\frac{24}{x^{3}}\rightarrow0) and (\frac{36}{x^{4}}\rightarrow0). So (\lim_{x\rightarrow\infty}\frac{\frac{1}{x}-\frac{7}{x^{2}}+\frac{16}{x^{3}}-\frac{12}{x^{4}}}{1-\frac{6}{x}+\frac{5}{x^{2}}+\frac{24}{x^{3}}-\frac{36}{x^{4}}}=\frac{0 - 0+0 - 0}{1-0 + 0+0 - 0}=0)

Answer:

0