find the limit, if it exists.\n51) $lim_{x\rightarrowinfty}\frac{x^{2}-4x + 14}{x^{3}+9x^{2}+5}$\n52)…

51)

Explanation:

Step1: Divide by highest - power of x in denominator

Divide both numerator and denominator by $x^{3}$. The function $\frac{x^{2}-4x + 14}{x^{3}+9x^{2}+5}$ becomes $\frac{\frac{x^{2}}{x^{3}}-\frac{4x}{x^{3}}+\frac{14}{x^{3}}}{\frac{x^{3}}{x^{3}}+\frac{9x^{2}}{x^{3}}+\frac{5}{x^{3}}}=\frac{\frac{1}{x}-\frac{4}{x^{2}}+\frac{14}{x^{3}}}{1 + \frac{9}{x}+\frac{5}{x^{3}}}$

Step2: Apply limit as $x\to\infty$

As $x\to\infty$, $\lim_{x\to\infty}\frac{1}{x}=0$, $\lim_{x\to\infty}\frac{4}{x^{2}} = 0$, $\lim_{x\to\infty}\frac{14}{x^{3}}=0$, $\lim_{x\to\infty}\frac{9}{x}=0$ and $\lim_{x\to\infty}\frac{5}{x^{3}}=0$. So, $\lim_{x\to\infty}\frac{\frac{1}{x}-\frac{4}{x^{2}}+\frac{14}{x^{3}}}{1+\frac{9}{x}+\frac{5}{x^{3}}}=\frac{0 - 0+0}{1 + 0+0}=0$

52)

Explanation:

Step1: Divide by highest - power of x in denominator

Divide both numerator and denominator by $x^{2}$. The function $\frac{-8x^{2}+9x + 5}{-15x^{2}+7x + 14}$ becomes $\frac{-8+\frac{9}{x}+\frac{5}{x^{2}}}{-15+\frac{7}{x}+\frac{14}{x^{2}}}$

Step2: Apply limit as $x\to-\infty$

As $x\to-\infty$, $\lim_{x\to-\infty}\frac{9}{x}=0$, $\lim_{x\to-\infty}\frac{5}{x^{2}} = 0$, $\lim_{x\to-\infty}\frac{7}{x}=0$ and $\lim_{x\to-\infty}\frac{14}{x^{2}}=0$. So, $\lim_{x\to-\infty}\frac{-8+\frac{9}{x}+\frac{5}{x^{2}}}{-15+\frac{7}{x}+\frac{14}{x^{2}}}=\frac{-8 + 0+0}{-15+0+0}=\frac{8}{15}$

53)

Explanation:

Step1: Divide by highest - power of x in denominator

Divide both numerator and denominator by $x$. The function $\frac{3x + 1}{11x-7}$ becomes $\frac{3+\frac{1}{x}}{11-\frac{7}{x}}$

Step2: Apply limit as $x\to\infty$

As $x\to\infty$, $\lim_{x\to\infty}\frac{1}{x}=0$ and $\lim_{x\to\infty}\frac{7}{x}=0$. So, $\lim_{x\to\infty}\frac{3+\frac{1}{x}}{11-\frac{7}{x}}=\frac{3 + 0}{11-0}=\frac{3}{11}$

54)

Explanation:

Step1: Divide by highest - power of x in denominator

Divide both numerator and denominator by $x^{3}$. The function $\frac{7x^{3}-3x^{2}+3x}{-x^{3}-2x + 7}$ becomes $\frac{7-\frac{3}{x}+\frac{3}{x^{2}}}{-1-\frac{2}{x^{2}}+\frac{7}{x^{3}}}$

Step2: Apply limit as $x\to\infty$

As $x\to\infty$, $\lim_{x\to\infty}\frac{3}{x}=0$, $\lim_{x\to\infty}\frac{3}{x^{2}} = 0$, $\lim_{x\to\infty}\frac{2}{x^{2}}=0$ and $\lim_{x\to\infty}\frac{7}{x^{3}}=0$. So, $\lim_{x\to\infty}\frac{7-\frac{3}{x}+\frac{3}{x^{2}}}{-1-\frac{2}{x^{2}}+\frac{7}{x^{3}}}=\frac{7-0 + 0}{-1-0+0}=-7$

Answer:

51. $0$
52. $\frac{8}{15}$
53. $\frac{3}{11}$
54. $-7$