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 numerator and denominator by $x^{3}$: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{x^{2}-4x + 14}{x^{3}+9x^{2}+5}&=\lim_{x\rightarrow\infty}\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}}}\ &=\lim_{x\rightarrow\infty}\frac{\frac{1}{x}-\frac{4}{x^{2}}+\frac{14}{x^{3}}}{1 + \frac{9}{x}+\frac{5}{x^{3}}} \end{align*} ]

Step2: Use limit rules for infinity

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{1}{x}=0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{3}}=0$. [ \begin{align*} \lim_{x\rightarrow\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 \end{align*} ]

Answer:

$0$

52)

Explanation:

Step1: Divide by highest - power of x in denominator

Divide numerator and denominator by $x^{2}$: [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{-8x^{2}+9x + 5}{-15x^{2}+7x + 14}&=\lim_{x\rightarrow-\infty}\frac{\frac{-8x^{2}}{x^{2}}+\frac{9x}{x^{2}}+\frac{5}{x^{2}}}{\frac{-15x^{2}}{x^{2}}+\frac{7x}{x^{2}}+\frac{14}{x^{2}}}\ &=\lim_{x\rightarrow-\infty}\frac{-8+\frac{9}{x}+\frac{5}{x^{2}}}{-15+\frac{7}{x}+\frac{14}{x^{2}}} \end{align*} ]

Step2: Use limit rules for infinity

As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{1}{x}=0$, $\lim_{x\rightarrow-\infty}\frac{1}{x^{2}} = 0$. [ \begin{align*} \lim_{x\rightarrow-\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} \end{align*} ]

Answer:

$\frac{8}{15}$

53)

Explanation:

Step1: Divide by highest - power of x in denominator

Divide numerator and denominator by $x$: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{3x + 1}{11x-7}&=\lim_{x\rightarrow\infty}\frac{\frac{3x}{x}+\frac{1}{x}}{\frac{11x}{x}-\frac{7}{x}}\ &=\lim_{x\rightarrow\infty}\frac{3+\frac{1}{x}}{11-\frac{7}{x}} \end{align*} ]

Step2: Use limit rules for infinity

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{1}{x}=0$. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{3+\frac{1}{x}}{11-\frac{7}{x}}&=\frac{3 + 0}{11-0}\ &=\frac{3}{11} \end{align*} ]

Answer:

$\frac{3}{11}$

54)

Explanation:

Step1: Divide by highest - power of x in denominator

Divide numerator and denominator by $x^{3}$: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{7x^{3}-3x^{2}+3x}{-x^{3}-2x + 7}&=\lim_{x\rightarrow\infty}\frac{\frac{7x^{3}}{x^{3}}-\frac{3x^{2}}{x^{3}}+\frac{3x}{x^{3}}}{\frac{-x^{3}}{x^{3}}-\frac{2x}{x^{3}}+\frac{7}{x^{3}}}\ &=\lim_{x\rightarrow\infty}\frac{7-\frac{3}{x}+\frac{3}{x^{2}}}{-1-\frac{2}{x^{2}}+\frac{7}{x^{3}}} \end{align*} ]

Step2: Use limit rules for infinity

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{1}{x}=0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$, $\lim_{x\rightarrow\infty}\frac{1}{x^{3}}=0$. [ \begin{align*} \lim_{x\rightarrow\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 \end{align*} ]

Answer:

$-7$