Answer
$$\lim_{x\to\infty}\frac{1}{x^2-7x+1}=0$$
Work Step by Step
*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\times\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$
$$A=\lim_{x\to\infty}\frac{1}{x^2-7x+1}$$
Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x^2$ in this case:
$$A=\lim_{x\to\infty}\frac{\frac{1}{x^2}}{1-\frac{7}{x}+\frac{1}{x^2}}$$
$$A=\frac{0}{1-0+0}=\frac{0}{1}=0$$
