Question

find the limit, if it exists.

1. $lim_{x

ightarrowinfty}\frac{x^{2}-4x + 14}{x^{3}+9x^{2}+5}$

1. $lim_{x

ightarrow-infty}\frac{-8x^{2}+9x + 5}{-15x^{2}+7x + 14}$

1. $lim_{x

ightarrowinfty}\frac{3x + 1}{11x-7}$

1. $lim_{x

ightarrowinfty}\frac{7x^{3}-3x^{2}+3x}{-x^{3}-2x + 7}$

Explanation:

51)

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)

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)

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)

Answer:

1. $0$
2. $\frac{8}{15}$
3. $\frac{3}{11}$
4. $-7$