Question

1. calculate the following limits at infinity. show all of your steps.

(a) $lim_{x
ightarrowinfty}\frac{3x^{2}-2x + 1}{2x^{2}-x + 1}$
(b) $lim_{x
ightarrow-infty}\frac{2x^{2}+1}{x + 2}$

Explanation:

Step1: Divide numerator and denominator by highest - power of x

For part (a), divide both the numerator and denominator of $\frac{3x^{2}-2x + 1}{2x^{2}-x + 1}$ by $x^{2}$:
$\lim_{x
ightarrow\infty}\frac{3x^{2}-2x + 1}{2x^{2}-x + 1}=\lim_{x
ightarrow\infty}\frac{3-\frac{2}{x}+\frac{1}{x^{2}}}{2-\frac{1}{x}+\frac{1}{x^{2}}}$

Step2: Use limit properties

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

For part (b), divide both the numerator and denominator of $\frac{2x^{2}+1}{x + 2}$ by $x$ (since the highest - power of $x$ in the denominator is 1).
$\lim_{x
ightarrow-\infty}\frac{2x^{2}+1}{x + 2}=\lim_{x
ightarrow-\infty}\frac{2x+\frac{1}{x}}{1+\frac{2}{x}}$
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{1}{x}=0$.
$\lim_{x
ightarrow-\infty}\frac{2x+\frac{1}{x}}{1+\frac{2}{x}}$, and as $x
ightarrow-\infty$, the term $2x
ightarrow-\infty$ and the terms $\frac{1}{x}
ightarrow0$ and $\frac{2}{x}
ightarrow0$. So, $\lim_{x
ightarrow-\infty}\frac{2x+\frac{1}{x}}{1+\frac{2}{x}}=-\infty$

Answer:

(a) $\frac{3}{2}$
(b) $-\infty$