Question

1. $lim_{x \to infty} \frac{3x^{2}+x^{2}-x - 1}{x^{3}+5x + 2}$

Explanation:

Step1: Divide by highest - power of x

Divide both the numerator and denominator by $x^{3}$ since the highest - power of $x$ in the denominator is $x^{3}$.
$\lim_{x
ightarrow\infty}\frac{3x^{2}+x^{2}-x - 1}{x^{3}+5x + 2}=\lim_{x
ightarrow\infty}\frac{\frac{3x^{2}}{x^{3}}+\frac{x^{2}}{x^{3}}-\frac{x}{x^{3}}-\frac{1}{x^{3}}}{\frac{x^{3}}{x^{3}}+\frac{5x}{x^{3}}+\frac{2}{x^{3}}}$

Step2: Simplify each term

Simplify the fractions in the above expression.
$\lim_{x
ightarrow\infty}\frac{\frac{3}{x}+\frac{1}{x}-\frac{1}{x^{2}}-\frac{1}{x^{3}}}{1+\frac{5}{x^{2}}+\frac{2}{x^{3}}}$

Step3: Use limit properties

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$.
$\frac{\lim_{x
ightarrow\infty}\frac{3}{x}+\lim_{x
ightarrow\infty}\frac{1}{x}-\lim_{x
ightarrow\infty}\frac{1}{x^{2}}-\lim_{x
ightarrow\infty}\frac{1}{x^{3}}}{\lim_{x
ightarrow\infty}1+\lim_{x
ightarrow\infty}\frac{5}{x^{2}}+\lim_{x
ightarrow\infty}\frac{2}{x^{3}}}=\frac{0 + 0-0 - 0}{1+0 + 0}$

Answer:

$0$